Specific Heat

Q: What is specific heat capacity?

Specific heat capacity is the heat needed to raise the temperature of one kilogram of a substance by one kelvin. Its SI unit is J kg⁻¹ K⁻¹. It is a property of the material: water has 4186, aluminium 900 and copper about 386.

Q: What is the difference between specific heat and heat capacity?

Specific heat is per unit mass and belongs to the material. Heat capacity is for the whole body, C = ms, so it also depends on how much material there is. A bucket and a cup of water have the same specific heat but different heat capacities.

Q: Why does water have a high specific heat and why does it matter?

Water molecules are held by hydrogen bonds, and much of the energy supplied goes into these bonds and molecular motions before the temperature rises. So water heats and cools slowly, which makes it an excellent coolant and keeps coastal climates mild.

Q: What is the mechanical equivalent of heat?

It is the number of joules of work that produce the same heating as one calorie of heat: J = 4.186 J cal⁻¹. Joule found it with a paddle wheel churning water. It is a conversion factor between units, not a physical constant.

Q: What is water equivalent?

Water equivalent is the mass of water that needs the same heat as a given body for the same rise in temperature. It equals W = ms/sw; in calorie units it is simply ms, in grams.

Q: Why is the specific heat of a gas at constant pressure greater than at constant volume?

At constant volume all the heat supplied raises the internal energy. At constant pressure the gas also expands and does work on its surroundings, so extra heat is needed for the same temperature rise. For an ideal gas CP - CV = R.

Q: How is specific heat asked in NEET?

NEET usually asks one direct question: heat needed for a temperature change, the ratio of temperature rises of two bodies, water equivalent, or conversion between temperature scales. Remember Q = ms ΔT and that a Celsius degree equals a kelvin.

Q: What type of specific heat questions come in JEE Main?

JEE Main mixes specific heat with energy conversion (falling bodies, bullets, heaters), temperature-dependent specific heat that must be integrated, faulty thermometer scales, and molar heat capacities of gases in different processes.

PhysicsThermal Properties Of MatterFor JEE aspirants

Specific heat is the heat needed to raise the temperature of unit mass of a substance by one degree: $s = \frac{Q}{m Δ T}$ . It tells you how "thermally stubborn" a material is: water, with the largest specific heat of common substances, warms and cools slowly. This page builds the language of heat (calorie, joule, mechanical equivalent, temperature scales) and then specific heat, heat capacity, molar specific heat and water equivalent, which are asked directly in NEET and JEE Main and are used in every calorimetry problem.

On this page1Heat and temperature2Temperature scales3Mechanical equivalent4Specific heat5Heat capacity6Molar specific heat7Water equivalent8Special cases

Key Formulas - Quick Reference

★ Must learn $Q = m s Δ T$ (constant $s$ ); $Q = m \int_{T_{1}}^{T_{2}} s d T$ when $s$ varies
★ Must learn $\frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273.15}{100}$ ; a change of $1^{\circ} C = 1 K = 1. 8^{\circ} F$
Mechanical equivalent of heat: $W = J H$ , $J = 4.186 J cal^{- 1}$ (1 cal $= 4.186 J$ )
★ Must learnSpecific heat of water $= 1 cal g^{- 1}^{\circ} C^{- 1} = 4186 J kg^{- 1} K^{- 1}$ ; ice and steam $\approx 0.5 cal g^{- 1}^{\circ} C^{- 1}$
Heat capacity (thermal capacity) $C = m s = \frac{Q}{Δ T}$ , unit $J K^{- 1}$
Molar specific heat $C_{m} = M s = \frac{1}{n} \frac{Δ Q}{Δ T}$ , unit $J mol^{- 1} K^{- 1}$ ; gases: $C_{P} - C_{V} = R$
★ Must learnWater equivalent $W = \frac{m s}{s _{w}}$ ( $= m s$ in calorie units), unit kg or g
Solids (Dulong-Petit): $C_{m} \approx 3 R \approx 25 J mol^{- 1} K^{- 1}$ at room temperature

1. Heat, Temperature and Internal Energy

★ Must learn

Heat is the energy transferred between two bodies (or two parts of a body) because of a temperature difference. It is energy in transit: once it arrives it becomes part of the internal energy of the receiving body. So "heat in a body" is meaningless; a body has internal energy.

Temperature is the property that decides the direction of heat flow: heat flows from higher to lower temperature, never the reverse on its own. When two bodies in contact stop exchanging heat they are in thermal equilibrium and have the same temperature. The zeroth law of thermodynamics (two bodies each in equilibrium with a third are in equilibrium with each other) is what lets a thermometer measure temperature.

Figure 1: Heat is the energy transferred because of a temperature difference. It always flows from the hotter body to the colder one and stops when both reach the same temperature (thermal equilibrium).

Heat	Temperature
Energy in transit due to a temperature difference	Degree of hotness; decides the direction of heat flow
SI unit joule (J); also calorie (cal)	SI unit kelvin (K); also °C, °F
Depends on mass, material and temperature change	Does not depend on the amount of matter
Microscopic view: transfer of random kinetic energy	Microscopic view: measure of average random kinetic energy of molecules

When a body is heated, its molecules move (or vibrate) faster: their average kinetic energy, and so the temperature, rises.

2. Measuring Temperature: Scales

A thermometer uses a property that changes steadily with temperature (length of a mercury column, pressure of a gas at constant volume, resistance of a wire). Two fixed points are chosen: the ice point (melting of pure ice at 1 atm) and the steam point (boiling of pure water at 1 atm).

Scale	Ice point	Steam point	Divisions between them
Celsius	$0^{\circ} C$	$10 0^{\circ} C$	100
Fahrenheit	$3 2^{\circ} F$	$21 2^{\circ} F$	180
Kelvin (absolute)	$273.15 K$	$373.15 K$	100

Figure 2: The same temperatures on the three scales. Between the ice and steam points there are 100 Celsius divisions, 180 Fahrenheit divisions and 100 kelvin, so

\frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273.15}{100}

★ Must learn

For any temperature, the fraction of the way from ice point to steam point is the same on every scale:

\frac{C - 0}{100} = \frac{F - 32}{212 - 32} = \frac{K - 273.15}{100}

General rule for any linear scale X with lower fixed point $L$ and upper fixed point $U$ : $\frac{X - L}{U - L}$ is the same on all scales.

Figure 3:

F = \frac{9}{5} C + 32

is a straight line (slope

1.8

, intercept

32

). It meets the dashed line

F = C

- 4 0^{\circ}

, the only temperature that reads the same on both scales. A change of

1^{\circ} C

equals a change of

1 K

and of

1. 8^{\circ} F

Absolute temperature. A constant-volume gas thermometer shows pressure falling linearly with temperature; extended, every gas line reaches $p = 0$ at the same temperature, $- 273.1 5^{\circ} C$ . This absolute zero is the zero of the Kelvin scale: $T (K) = t (^{\circ} C) + 273.15$ . For an ideal gas, $p V = μ R T$ with $T$ in kelvin. The triple point of water, $273.16 K$ ( $0.0 1^{\circ} C$ ), is the modern fixed point of the Kelvin scale.

Exam Trick

Differences convert without the offset. A temperature converts with $+ 32$ or $+ 273.15$ ; a temperature difference does not: $Δ T (K) = Δ t (^{\circ} C)$ and $Δ F = 1.8 Δ C$ . That is why specific heat has the same number in $J kg^{- 1} K^{- 1}$ and $J kg^{- 1}^{\circ} C^{- 1}$ . For a faulty thermometer, use $\frac{reading - lower mark}{upper mark - lower mark}$ .

3. Mechanical Equivalent of Heat

Heat was once thought to be a fluid ("caloric") and was measured in calories. Joule showed that doing mechanical work on a system raises its temperature exactly as heat does, with a fixed exchange rate. If work $W$ produces the same effect as heat $H$ :

W = J H, J = 4.186 J cal^{- 1}

$J$ is not a physical constant but a conversion factor between two units of the same quantity (energy). In SI, heat is measured in joules directly and $J$ disappears.

1 calorie is the heat needed to raise the temperature of $1 g$ of water from $14. 5^{\circ} C$ to $15. 5^{\circ} C$ at 1 atm. $1 cal = 4.186 J$ and $1 kcal = 1000 cal = 4186 J$ (the food "Calorie" is 1 kcal).

Figure 4: Joule's experiment. The work

W = 2 M g h

done by the falling weights churns the water and raises its temperature exactly as if heat

H

had been supplied:

W = J H

with

J = 4.186 J cal^{- 1}

Key idea

Work and heat are two routes for the same energy:

W = J H

. In SI units both are joules.

4. Specific Heat Capacity

Experiments show that the heat $Δ Q$ gained or lost by a body is proportional to its mass $m$ and to its change of temperature $Δ T$ :

Δ Q \propto m Δ T \Rightarrow Δ Q = m s Δ T

★ Must learn

The specific heat capacity $s$ of a substance is the heat needed to raise the temperature of unit mass by one degree:

s = \frac{1}{m} \frac{Δ Q}{Δ T}

SI unit $J kg^{- 1} K^{- 1}$ ; CGS unit $cal g^{- 1}^{\circ} C^{- 1}$ ; dimensions $[L^{2} T^{- 2} K^{- 1}]$ . It depends on the material and its state (and slightly on temperature), not on the size of the body.

If $s$ depends on temperature, add up small steps: $d Q = m s d T$ , so $Q = m \int_{T_{1}}^{T_{2}} s d T$ .

Substance	$s$ ( $J kg^{- 1} K^{- 1}$ )	Substance	$s$ ( $J kg^{- 1} K^{- 1}$ )
Water	4186	Aluminium	900
Ice	2060	Glass	840
Kerosene	2118	Copper	386.4
Sea water	3900	Silver	236.1
Edible oil	1965	Lead	127.7

Figure 5: For

1 kg

of each (specific heats in

J kg^{- 1} K^{- 1}

Δ T = \frac{Q}{m s}

: the slope is

\frac{1}{m s}

. The smaller the specific heat, the steeper the line. Water heats up (and cools down) the slowest, which is why it is used as a coolant.

Figure 6: Specific heat capacities in

J kg^{- 1} K^{- 1}

at room temperature (NCERT values). Water's is the largest of common substances; metals have small specific heats, so they warm up quickly.

Why water's high specific heat matters. Water absorbs a lot of heat for a small rise in temperature, so it is used as a coolant in car radiators and power plants, in hot-water bags, and it keeps coastal climates mild (the sea warms and cools slowly; this also drives sea and land breezes).

4.1 Specific heat of water varies slightly

The specific heat of water is not exactly constant. Between $0^{\circ} C$ and $10 0^{\circ} C$ it changes by less than 1%, with a shallow minimum near $3 5^{\circ} C$ . For problems take $1 cal g^{- 1}^{\circ} C^{- 1}$ ( $4186$ or $4200 J kg^{- 1} K^{- 1}$ as the question says).

Figure 7: The specific heat of water changes by less than 1% between

0^{\circ} C

and

10 0^{\circ} C

, with a shallow minimum near

3 5^{\circ} C

(curve from steam-table data). That is why the calorie is defined for the interval

14. 5^{\circ} C

15. 5^{\circ} C

Quick Recall: tap to check

A body of mass

2 kg

has

s = 400 J kg^{- 1} K^{- 1}

. What is its heat capacity?

C = m s = 800 J K^{- 1}

Is the value of specific heat different in

J kg^{- 1} K^{- 1}

and

J kg^{- 1}^{\circ} C^{- 1}

No. A 1 K change equals a 1 °C change.

Why is the calorie defined between 14.5 °C and 15.5 °C?

Because the specific heat of water varies slightly with temperature.

5. Heat Capacity, Molar Specific Heat and Water Equivalent

5.1 Heat capacity (thermal capacity)

The heat capacity of a body is the heat needed to raise the temperature of the whole body by one degree:

C = \frac{Δ Q}{Δ T} = m s

Unit $J K^{- 1}$ (CGS: $cal^{\circ} C^{- 1}$ ). It depends on both the material and the mass, so a bucket of water has a larger heat capacity than a cup of water, though both have the same specific heat.

Specific heat

s

Per unit mass. Property of the material. Unit $J kg^{- 1} K^{- 1}$ . Same for a cup or a bucket of water.

Heat capacity

C = m s

For the whole body. Depends on mass too. Unit $J K^{- 1}$ . A bucket of water has a larger $C$ than a cup.

5.2 Molar specific heat

Per mole instead of per kilogram: $C_{m} = \frac{1}{n} \frac{Δ Q}{Δ T} = M s$ , where $M$ is the molar mass in $kg mol^{- 1}$ . Unit $J mol^{- 1} K^{- 1}$ .

Solids: most metals have $C_{m} \approx 3 R \approx 25 J mol^{- 1} K^{- 1}$ at room temperature (law of Dulong and Petit). Copper: $386.4 \times 0.0635 \approx 24.5 J mol^{- 1} K^{- 1}$ . At low temperatures $C_{m}$ falls towards zero.
Gases: the heat needed depends on the process. At constant volume $C_{V}$ ; at constant pressure $C_{P}$ . $C_{P} > C_{V}$ because at constant pressure part of the heat does work in expansion: $C_{P} - C_{V} = R$ (Mayer's relation) for an ideal gas.

Gas type	$C_{V}$	$C_{P}$	$γ = C_{P} / C_{V}$
Monatomic (He, Ar)	$\frac{3}{2} R$	$\frac{5}{2} R$	$\frac{5}{3} \approx 1.67$
Diatomic ( $O_{2}$ , $N_{2}$ ), room temperature	$\frac{5}{2} R$	$\frac{7}{2} R$	$\frac{7}{5} = 1.4$
Non-linear polyatomic ( $C H_{4}$ ), no vibration	$3 R$	$4 R$	$\frac{4}{3} \approx 1.33$

5.3 Water equivalent

The water equivalent $W$ of a body is the mass of water that would need the same heat as the body for the same rise in temperature:

m s Δ T = W s_{w} Δ T \Rightarrow W = \frac{m s}{s _{w}}

In calorie units $s_{w} = 1$ , so $W = m s$ numerically (in grams). Its unit is that of mass. In calorimetry, a calorimeter of water equivalent $W$ is simply treated as $W$ extra grams of water.

Figure 8: The water equivalent

W

of a body is the mass of water that needs the same heat for the same rise in temperature:

W = \frac{m s}{s _{w}}

, i.e.

W = m s

with

s

cal g^{- 1}^{\circ} C^{- 1}

Key idea

s

belongs to the material,

C = m s

to the body,

C_{m} = M s

to a mole, and

W = m s / s_{w}

turns any body into an equivalent mass of water.

6. Special Cases and the Right Formula

The ratio $\frac{Q}{m Δ T}$ can be worked out for any process, and in some processes it gives strange values. These describe the process, not a new property of the material:

Situation	$Q$	$Δ T$	$\frac{Q}{m Δ T}$
Melting or boiling (phase change)	supplied	0	infinite: heat goes into latent heat
Adiabatic process (no heat exchanged, e.g. quick compression of a gas)	0	not zero	zero
Saturated water vapour kept saturated while heated (for information only)	must be removed	positive	negative
Ordinary heating at constant pressure	supplied	positive	the tabulated specific heat

A liquid in a thermos flask that warms when shaken does not have zero specific heat. No heat enters, but work is done on it by the shaking: its internal energy and temperature rise through work, $W = J H$ . Its specific heat is still the tabulated value.

Figure 9: Choosing the right heat formula. Most problems end at

Q = m s Δ T

; watch for a phase change inside the range, a gas (use molar heat capacities) or a temperature-dependent

s

(integrate).

JEE Advanced

For a gas the molar heat capacity depends on the path: isothermal process $C = \infty$ , adiabatic $C = 0$ , isobaric $C_{P}$ , isochoric $C_{V}$ , and for a polytropic process $p V^{x} = constant$ : $C = C_{V} + \frac{R}{1 - x}$ , which is negative when $1 < x < γ$ . When $s$ is given as a function of temperature, e.g. $s = a + b T^{2}$ , always integrate: $Q = m [a Δ T + \frac{b}{3} (T_{2}^{3} - T_{1}^{3})]$ .

Figure 10: Mind map of this concept: revise from the centre outwards and recall each branch's formula and unit before checking.

7. Solved Examples

Solved Example 1

Find the heat required to raise the temperature of

1 kg

of water by

2 0^{\circ} C

Solution:

$Q = m s Δ T = (1000 g) (1 cal g^{- 1}^{\circ} C^{- 1}) (2 0^{\circ} C) = 20000 cal$ .

Answer: $Q = 20 kcal \approx 83.7 kJ$ . (Here $s = 1 cal g^{- 1}^{\circ} C^{- 1} = 1 kcal kg^{- 1}^{\circ} C^{- 1}$ .)

Solved Example 2

An iron block of mass

2 kg

falls from a height of

10 m

. On hitting the ground it loses

25%

of the energy to the surroundings and the rest heats the block. Find the rise in temperature of the block. (

s_{iron} = 470 J kg^{- 1}^{\circ} C^{- 1}

g = 10 m s^{- 2}

)

Solution:

Energy lost to the surroundings is $25%$ , so the block keeps $75%$ : $m s Δ θ = \frac{3}{4} m g h$ .

$Δ θ = \frac{3 g h}{4 s} = \frac{3 ( 10 ) ( 10 )}{4 ( 470 )}$ . The mass cancels.

Answer: $Δ θ \approx 0.1 6^{\circ} C$ .

Solved Example 3

(a) Convert normal body temperature

98. 6^{\circ} F

to Celsius and kelvin. (b) At what temperature do the Kelvin and Fahrenheit scales give the same reading?

Solution:

(a) $\frac{C}{100} = \frac{98.6 - 32}{180}$ gives $C = 3 7^{\circ} C$ , and $K = 37 + 273.15 = 310.15 K$ .

(b) Put $F = K = x$ : $\frac{x - 32}{180} = \frac{x - 273.15}{100}$ , so $100 x - 3200 = 180 x - 49167$ , giving $80 x = 45967$ .

Answer: (a) $3 7^{\circ} C = 310.15 K$ ; (b) $x \approx 574.6$ , i.e. $574.6 K = 574. 6^{\circ} F$ .

Solved Example 4

A faulty thermometer reads

5^{\circ}

in melting ice and

9 5^{\circ}

in steam at 1 atm. What is the true Celsius temperature when it reads

5 0^{\circ}

Solution:

$\frac{C - 0}{100} = \frac{50 - 5}{95 - 5} = \frac{45}{90}$ .

Answer: $C = 5 0^{\circ} C$ . (This faulty scale agrees with the true one only at $5 0^{\circ}$ .)

Solved Example 5

Water falls from a height of

100 m

. If all its potential energy is converted into heat that stays in the water, find the rise in temperature. (

s = 4186 J kg^{- 1} K^{- 1}

g = 9.8 m s^{- 2}

)

Solution:

$m g h = m s Δ T \Rightarrow Δ T = \frac{g h}{s} = \frac{9.8 \times 100}{4186}$ .

Answer: $Δ T \approx 0.23 K$ , independent of the mass.

Solved Example 6

A lead bullet moving at

200 m s^{- 1}

stops in a wooden block. If half its kinetic energy heats the bullet, find its rise in temperature. (

s_{lead} = 127.7 J kg^{- 1} K^{- 1}

)

Solution:

$m s Δ T = \frac{1}{2} \cdot \frac{1}{2} m v^{2} \Rightarrow Δ T = \frac{v ^{2}}{4 s} = \frac{( 200 ) ^{2}}{4 ( 127.7 )}$ .

Answer: $Δ T \approx 7 8^{\circ} C$ .

Solved Example 7

An electric kettle of power

1 kW

heats

2 kg

of water from

2 0^{\circ} C

10 0^{\circ} C

. Ignoring losses and the kettle's own heat capacity, how long does it take? (

s = 4186 J kg^{- 1} K^{- 1}

)

Solution:

$Q = m s Δ T = (2) (4186) (80) = 6.70 \times 1 0^{5} J$ .

$t = \frac{Q}{P} = \frac{6.70 \times 1 0 ^{5}}{1000}$ .

Answer: $t \approx 670 s \approx 11.2 min$ .

Solved Example 8

The specific heat of a substance varies as

s = (0.3 + 0.002 T) cal g^{- 1}^{\circ} C^{- 1}

, with

T

^{\circ} C

. Find the heat needed to raise

100 g

of it from

0^{\circ} C

10 0^{\circ} C

Solution:

$Q = m \int_{0}^{100} (0.3 + 0.002 T) d T = 100 [0.3 T + 0.001 T^{2}]_{0}^{100} = 100 (30 + 10)$ .

Answer: $Q = 4000 cal$ . (Using the value of $s$ at the start, $0.3$ , would give only $3000 cal$ .)

Solved Example 9

A copper calorimeter has mass

200 g

. Find its heat capacity and water equivalent. (

s_{Cu} = 0.09 cal g^{- 1}^{\circ} C^{- 1}

)

Solution:

Heat capacity $C = m s = (200) (0.09) = 18 cal^{\circ} C^{- 1}$ .

Water equivalent $W = \frac{m s}{s _{w}} = \frac{18}{1}$ .

Answer: $C = 18 cal^{\circ} C^{- 1}$ ( $\approx 75 J K^{- 1}$ ); $W = 18 g$ .

Solved Example 10

Two bodies A and B have masses in the ratio

1 : 2

and specific heats in the ratio

2 : 1

. The same amount of heat is given to each. The ratio of their temperature rises

Δ T_{A} : Δ T_{B}

is
(A)

1 : 4

(B)

1 : 1

(C)

4 : 1

(D)

2 : 1

Solution:

Answer: (B). $Δ T = \frac{Q}{m s}$ and $m_{A} s_{A} = (1) (2) = m_{B} s_{B} = (2) (1)$ : equal heat capacities give equal rises.

Solved Example 11

The molar mass of copper is

63.5 g mol^{- 1}

and its specific heat is

386.4 J kg^{- 1} K^{- 1}

. Find its molar specific heat and compare it with

3 R

Solution:

$C_{m} = M s = (0.0635) (386.4) = 24.5 J mol^{- 1} K^{- 1}$ ; $3 R = 3 (8.314) = 24.9 J mol^{- 1} K^{- 1}$ .

Answer: $C_{m} \approx 24.5 J mol^{- 1} K^{- 1}$ , within 2% of $3 R$ (Dulong-Petit law).

Practice Questions

Convert $- 4 0^{\circ} C$ and $77 K$ to Fahrenheit.Answer: $- 4 0^{\circ} F$ ; $- 321. 1^{\circ} F$
How much heat raises $500 g$ of aluminium ( $s = 900 J kg^{- 1} K^{- 1}$ ) from $2 0^{\circ} C$ to $8 0^{\circ} C$ ?Answer: $27 kJ$
A $2 kW$ heater warms a $4 kg$ block by $25 K$ in $50 s$ . Find the specific heat of the block (no losses).Answer: $1000 J kg^{- 1} K^{- 1}$
A body of heat capacity $50 J K^{- 1}$ has mass $0.25 kg$ . Find its specific heat and water equivalent.Answer: $200 J kg^{- 1} K^{- 1}$ ; about $11.9 g$
What height must water fall so that its temperature rises by $1 K$ if all the energy stays in it? ( $g = 9.8 m s^{- 2}$ )Answer: about $427 m$
Why is $C_{P}$ of a gas larger than $C_{V}$ ?Answer: At constant pressure part of the heat does work as the gas expands
A thermometer reads $- 1 0^{\circ}$ at the ice point and $11 0^{\circ}$ at the steam point. What is the true temperature when it reads $7 0^{\circ}$ ?Answer: about $66. 7^{\circ} C$

Common Mistakes to Avoid

Watch out

Adding $273$ to a temperature difference. $Δ T$ in kelvin equals $Δ t$ in $^{\circ} C$ .
Using $s = 1$ for water in SI problems. It is $1 cal g^{- 1}^{\circ} C^{- 1}$ or $4186 J kg^{- 1} K^{- 1}$ ; keep units consistent (grams with calories, kilograms with joules).
Confusing specific heat (per kg, material property) with heat capacity (whole body, depends on mass).
Saying a body \"contains heat\". A body has internal energy; heat is only the energy transferred.
Using $Q = m s Δ T$ across a melting or boiling point. Split the range and add $m L$ at the phase change.
Taking $J = 4.186$ as a physical constant. It is only the number of joules in one calorie.
Using the initial value of a temperature-dependent $s$ instead of integrating $m \int s d T$ .
In energy-conversion problems, forgetting to use only the fraction of energy that actually becomes heat in the body.

Frequently Asked Questions

What is specific heat capacity?

Specific heat capacity is the heat needed to raise the temperature of one kilogram of a substance by one kelvin. Its SI unit is $J kg^{- 1} K^{- 1}$ . It is a property of the material: water has 4186, aluminium 900 and copper about 386.

What is the difference between specific heat and heat capacity?

Specific heat is per unit mass and belongs to the material. Heat capacity is for the whole body, $C = m s$ , so it also depends on how much material there is. A bucket and a cup of water have the same specific heat but different heat capacities.

Why does water have a high specific heat and why does it matter?

Water molecules are held by hydrogen bonds, and much of the energy supplied goes into these bonds and molecular motions before the temperature rises. So water heats and cools slowly, which makes it an excellent coolant and keeps coastal climates mild.

What is the mechanical equivalent of heat?

It is the number of joules of work that produce the same heating as one calorie of heat: $J = 4.186 J cal^{- 1}$ . Joule found it with a paddle wheel churning water. It is a conversion factor between units, not a physical constant.

What is water equivalent?

Water equivalent is the mass of water that needs the same heat as a given body for the same rise in temperature. It equals $W = \frac{m s}{s _{w}}$ ; in calorie units it is simply $m s$ , in grams.

Why is the specific heat of a gas at constant pressure greater than at constant volume?

At constant volume all the heat supplied raises the internal energy. At constant pressure the gas also expands and does work on its surroundings, so extra heat is needed for the same temperature rise. For an ideal gas $C_{P} - C_{V} = R$ .

How is specific heat asked in NEET?

NEET usually asks one direct question: heat needed for a temperature change, the ratio of temperature rises of two bodies, water equivalent, or conversion between temperature scales. Remember $Q = m s Δ T$ and that a Celsius degree equals a kelvin.

What type of specific heat questions come in JEE Main?

JEE Main mixes specific heat with energy conversion (falling bodies, bullets, heaters), temperature-dependent specific heat that must be integrated, faulty thermometer scales, and molar heat capacities of gases in different processes.

Previous year questions on Specific Heat

2 questions from past papers, each with a step-by-step solution.

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