Practice Specific Heat Capacity in Physics

Use these practice problems to test your method after reviewing the concept explanation and worked examples.

Worksheet PDF Free Answer-key PDF Family

Quick Recap

Specific heat capacity is the amount of energy needed to raise the temperature of 1 kilogram of a substance by 1 degree Celsius (equivalently, by 1 kelvin).

Some substances warm up quickly, while others need much more energy for the same temperature change.

Showing a random 20 of 50 problems.

Example 1

easy

True or false: a

5\text{ K}

rise and a

5^\circ\text{C}

rise mean different amounts of heat (for the same mass and substance).

Example 2

easy

Find the heat to warm

3 \text{ kg}

of aluminium (

c=900

) by

5^\circ\text{C}

Example 3

medium

0.2 \text{ kg}

of oil (

c=2000

) and

0.2 \text{ kg}

of water (

c=4200

) each receive

4200 \text{ J}

. Which warms more, and by how much more?

Example 4

easy

It takes

4200 \text{ J}

to raise

1 \text{ kg}

of water by

1^\circ\text{C}

. How much for

1 \text{ kg}

3^\circ\text{C}

Example 5

hard

0.1\text{ kg}

of iron (

c_{Fe} = 450

) at

400^\circ\text{C}

is quenched in

0.3\text{ kg}

of oil (

c_o = 2000

) at

25^\circ\text{C}

. Find

T_f

Example 6

medium

500 \text{ g}

of water (

c=4200

) is heated from

20^\circ\text{C}

80^\circ\text{C}

. Find the heat in kilojoules.

Example 7

easy

How much does

0.5 \text{ kg}

of water (

c=4200

) rise in temperature when given

21000 \text{ J}

Example 8

medium

2.5\text{ kg}

of water (

c = 4200

) cools from

80^\circ\text{C}

30^\circ\text{C}

. How much heat is released?

Example 9

easy

Find the heat to warm

0.25 \text{ kg}

of copper (

c=400

) by

40^\circ\text{C}

Example 10

challenge

A swimming pool holds

30\,000\text{ kg}

of water (

c = 4200

). A

20\text{ kW}

heater runs

8\text{ h/day}

with 10% heat loss. Find

\Delta T

per day.

Example 11

easy

What are the SI units of specific heat capacity?

Example 12

easy

How much heat warms

200\text{ g}

of water (

c = 4200\text{ J/(kg}\cdot{}^\circ\text{C)}

) from

20^\circ\text{C}

25^\circ\text{C}

Example 13

hard

2500\text{ W}

heater loses

300\text{ W}

to the air. How long to raise

3\text{ kg}

of water (

c = 4200

) by

40^\circ\text{C}

Example 14

medium

750\text{ W}

kettle heats

0.6\text{ kg}

of water (

c = 4200\text{ J/(kg}\cdot{}^\circ\text{C)}

) from

20^\circ\text{C}

100^\circ\text{C}

, ignoring losses. How long does it take?

Example 15

medium

Equal heat is given to

1 \text{ kg}

each of water (

c=4200

) and iron (

c=450

). The iron rises

40^\circ\text{C}

. By how much does the water rise?

Example 16

medium

Equal masses of water (

c = 4200

) and iron (

c = 450

) each gain

9000\text{ J}

. Which warms more, and what is the ratio of temperature rises (

\Delta T_{Fe}/\Delta T_{H_2O}

Example 17

hard

How much water (kg) at

15^\circ\text{C}

must be added to

2\text{ kg}

of water at

90^\circ\text{C}

to reach

50^\circ\text{C}

? Use

c = 4200

Example 18

challenge

1 \text{ kg}

of water (

c=4200

) at

30^\circ\text{C}

is mixed with

m

kg at

80^\circ\text{C}

to reach

50^\circ\text{C}

. Find

m

Example 19

easy

How much heat is needed to raise

0.25\text{ kg}

of aluminium (

c = 900\text{ J/(kg}\cdot{}^\circ\text{C)}

) by

20^\circ\text{C}

Example 20

medium

1800\text{ W}

kettle takes

3\text{ min}

to fully heat water with no losses. If the water is

1.0\text{ kg}

(

c = 4200

), find

\Delta T

← Back to Specific Heat Capacity Worked Examples Formula Explained

Related Concepts

Thermal Equilibrium Thermal Energy Temperature Ideal Gas Law