The equation is:

Q = mc_pΔT
Heat = (mass) (specific heat) (change in temperature)

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

Step 1: Organize the information Mass: m = 0.25 kg

Specific heat: c_p = 4.186 * 10³ J kg*°C

Change in temperature: ΔT = 2.5°C

Heat: Q = ? Close

Step 2: Equation Q = mc_pΔT Close

Step 3: Solve Q = (0.25)(4.186 * 10³)(2.5)

Q = 2616.25 J Close

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?

Step 1: Organize the information Mass = m = 0.15 kg

Specific heat: c_p = 460 J kg*°C

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

Change in temperature: ΔT = ?

T_i = 400°C

T_f = ? Close

Step 2: Equations Q = mc_pΔT

ΔT = Q/mc_p

ΔT = T_f – T_i

T_f = T_i + ΔT Close

Step 3: Solve ΔT = Q/mc_p

ΔT = (-24150)/(0.15)(460)

ΔT = -350C

T_f = T_i + ΔT

T_f = 400 + (-350)

T_f = 50°C Close

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?

Step 1: Organize all the information in the problem

quantity	variable	Aluminum	Water
mass	m	m_AL = 0.050kg	m_w = 0.15kg
specific heat	c_p	c_p-AL = 8.99 * 10² Joules per kilogram degree Celsius J kg*°C	c_p-w = 4.186 * 10³ Joules per kilogram degree Celsius J kg*°C
initial temperature	T_i	T_i-AL = ?	T_i-w = 21°C
final temperature	T_f	T_f-AL = 25°C	T_f-w = 25°C
change in temperature	ΔT	ΔT_AL = ?	ΔT_w = 25.0°C – 21.0°C = 4.0°
heat transferred	Q	Q_AL = ?	Q_W = ?

Close

Step 2: Equation Q = mc_pΔT Close

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. Close

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 Close

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 Close

Step 5: Find Ti 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 Close