Some materials heat up and cool down faster than others. The amount of energy required to raise the temperature of a 1 gram (or 1kg depending on units) of a substance by 1°C (or by 1 Kelvin) is called the specific heat (c_p). The specific heat of a material is a constant, but its numerical value depends on whether you are measuring your energy in Joules or calories, and whether you are measuring mass in grams or kilograms. For the sample problems in this lesson, you will be using Joules for energy, and kilograms for mass.

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Watch the following video showing how to solve a specific heat problem from start to finish.

Now let's try a few sample problems.

Sample 1

How much heat needs to be added to 0.25kg of water to raise it 2.5°C? The specific heat of water is 4.186 × 10³ Joules per kilogram degree Celsius J kg°C

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Step 1: Organize the information.

Mass: m = 0.25 kg

Specific heat: c_p = 4.186 × 10³ Joules per kilogram degree Celsius J kg×°C

Change in temperature: ΔT = 2.5°C

Heat: Q = ?

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Step 2: Equation

Q = mc_pΔT

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Step 3: Solve

Q = (0.25)(4.186 × 10³)(2.5)

Q = 2616.25 J

Sample 2

A 0.15 kg bar of iron (specific heat = 460 Joules per kilogram degree Celsius J kg°C that is originally at 400°C is set out to cool. Some time later, the bar has lost 24,150 Joules of heat. What is the new temperature of the bar?

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Step 1: Organize the information.

Mass: m = 0.15 kg

Specific heat: c_p = 460 Joules per kilogram degree Celsius J kg°C

Heat: Q = -24,150 J (negative because the bar lost that much heat)

Change in temperature: ΔT = ?

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Step 2: Equation

Q = mc_pΔT

ΔT = Q/mc_p

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Step 3: Solve

ΔT = Q/mc_p

ΔT = (-24150)/(0.15)(460)

ΔT = -350C

Sample 3

A bar of aluminum with a mass of 0.050 kg is at an unknown initial temperature. The bar of aluminum is then dropped into a container that contains 0.15 kg of water that is at an initial temperature of 21.0°C. The bar of aluminum is left in the water until they both reach a final temperature of 25.0°C. The specific heat for aluminum is 8.99 × 10² Joules per kilogram degree Celsius J kg°C , and the specific heat for water is 4.186 × 10³ Joules per kilogram degree Celsius J kg°C . What was the initial temperature of the aluminum bar?

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Step 1: Organize the information.

quantity	variable	Aluminum	Water
mass	m	mAL = 0.050kg	mw = 0.15kg
specific heat	cp	cp-AL = 8.99 × 102 Joules per kilogram degree Celsius J kg°C	cp-w = 4.186 × 103 Joules per kilogram degree Celsius J kg°C
initial temperature	Ti	Ti-AL = ?	Ti-w = 21°C
final temperature	Tf	Tf-AL = 25°C	Tf-w = 25°C
change in temperature	ΔT	ΔTAL = ?	ΔTw = 25.0°C – 21.0°C = 4.0°
heat transferred	Q	QAL = ?	QW = ?

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Step 2: Equation

Q = mc_pΔT

ΔT = Q/mc_p


ΔT = T_f – T_i

T_f = T_i + ΔT

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Hint

Since energy is conserved, the amount of heat energy lost by the aluminum bar will be equal to the amount of heat energy gained by the water.

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Step 3: Find the heat gained by the water.

Q = mc_pΔT

Q_W = (0.15)(4.186 × 10³)(4)

Q_W = 2511.6J

Q_AL = -2511.6J

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Step 4: Find ΔT for the aluminum.

Q = mcp × T

ΔT_AL = Q/mc_p

ΔT_AL = -2511.6/(.05) (8.99 × 10²)

ΔT_AL = -55.88°C

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Step 5: Find T_i for the aluminum.

ΔT_AL = T_f-AL – T_i-AL

T_i-AL = T_f-AL – ΔT_AL

T_i-AL = 25 – (-55.88)

T_i-AL = 80.88°C