Specific Heat Capacity Calculations

SCI 10: Weather Dynamics

Specific Heat Capacity Calculations: Part IThe amount of heat energy (Q) gained or lost by a substance is equal to the mass of the substance (m) multiplied by its specific heat capacity (C) multiplied by the change in temperature (final temperature - initial temperature)

Q = mC ΔT (ΔT = “change in temperature”) Note: ΔT = Tfinal – Tinitial à ΔT = Tf – Ti

So: Q = m x C x (Tf - Ti)

·  Specific Heat Capacity (C) of a substance is the amount of heat required to raise the temperature of 1g of the substance by 1oC.

·  The units of specific heat capacity are J/goC

Specific Heat Capacities of Some Substances
[C measured in J o/goC)]
Aluminium / C = 0.90 / water / C = 4.18
Carbon / C = 0.72 / ethanol (ethyl alcohol) / C = 2.44
Copper / C = 0.39 / sulfuric acid (liquid) / C = 1.42
Lead / C = 0.13 / sodium chloride solid / C = 0.85
Mercury / C = 0.14 / potassium hydroxide solid / C = 1.18

Practice Problems

1.  Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 56oC.

Q = mCΔT
Q = m x C x ΔT
Q = m x C x (Tf - Ti)
m = 250g
C = 4.18 J/goC (from table above)
Tf = 56oC
Ti = 20oC
Q = m x C x (Tf - Ti)
Q = 250g x 4.18 J/goC x (56oC - 20oC)
Q = 250g x 4.18 J/goC x (36 oC)
Q = 37 620 J

2.  Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25o to 60o.

Q = m x C x (Tf - Ti) OR Use Equation C = ___Q____
m(Tf - Ti)
Q = 204.75 J
m = 15g
Ti = 25oC
Tf = 60oC

204.75 J = 15g x C x (60 oC - 25 oC)
204.75 J = 15g x C x 35 oC
204.75 J = 525g oC x C
C = 204.75 J ÷ 525goC
C = 0.39 J/goC

3.  216 J of energy is required to raise the temperature of aluminium from 15o to 35oC. Calculate the mass of aluminium. (Specific Heat Capacity of aluminium is 0.90 J/goC).

Q = m x Cg x (Tf - Ti) OR Use Equation m = ___Q____
C(Tf - Ti)
Q = 216 J
C = 0.90 J/goC
Ti = 15oC
Tf = 35oC

216 J = m x 0.90 J/goC x (35oC - 15oC)
216 J = m x 0.90 J/goC x 20oC
216 J = m x 18 J/g
m = 216 J ÷ 18 J/g
m = 12g

4.  The initial temperature of 150g of ethanol was 22oC. What will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol?

Q = m x Cg x (Tf - Ti)
Q = 3240 J
m = 150g
C = 2.44 J/goC
Ti = 22oC

3240 J = 150g x 2.44 J/goC x (Tf - 22oC)
3240 J = 36644 J/oC (Tf - 22oC)
8.85 = Tf - 22oC
Tf = 30.9oC