Name: _________________________
General Science C
Date: ______________
Mr. Tiesler
Solutions to Specific Heat Problems I
1.) The specific heat of iron is 0.4494 J/g∙oC. How much heat is transferred when a 4.7 kg piece
of iron is cooled from 180 oC to 13 oC? Remember you must use the same units so you will have
to convert your mass to grams before you begin.
𝑐 = 0.4494 𝐽⁄𝑔 ∙ ℃
𝑚 = 4.7 𝑘𝑔 = 4,700 𝑔
∆𝑇 = 13 ℃ − 180 ℃ = −167 ℃
𝑄 =?
𝑄 = 𝑚𝑐∆𝑇
𝑄 = (4,700 𝑔)(0.4494 𝐽⁄𝑔 ∙ ℃)(−167 ℃) = −352,734 𝐽
2.) How many grams of water would require 92.048 kJ of heat to raise its temperature from 34.0
o
C to 100.0 oC? The specific heat of water is 4.18 J/g∙oC. (Remember to change units first)
𝑄 = 92.048 𝑘𝐽 = 92,048 𝐽
∆𝑇 = 100.0 ℃ − 34.0 ℃ = 66.0 ℃
𝑐 = 4.18 𝐽⁄𝑔 ∙ ℃
𝑚 =?
𝑄 = 𝑚𝑐∆𝑇
𝑚=
𝑄
92,048 𝐽
=
= 333 𝑔
𝑐∆𝑇 (4.18 𝐽⁄𝑔 ∙ ℃)(66.0 ℃)
3.) How many degrees would the temperature of a 450 g piece of iron increase if 7600 J of
energy are applied to it? (The specific heat of iron is 0.4494 J/g∙oC)
𝑚 = 450 𝑔
𝑄 = 7600 𝐽
𝑐 = 0.4494 𝐽⁄𝑔 ∙ ℃
∆𝑇 =?
𝑄 = 𝑚𝑐∆𝑇
∆𝑇 =
𝑄
7600 𝐽
=
= 37.6 ℃
𝑚𝑐 (450 𝑔)(0.4494 𝐽⁄𝑔 ∙ ℃)
4.) How much change in temperature would the addition of 35 000 Joules of heat have on a
538.0 gram sample of copper? The specific heat of copper is 0.385 J/g∙oC.
𝑄 = 35,000 𝐽
𝑚 = 538.0 𝑔
𝑐 = 0.385 𝐽⁄𝑔 ∙ ℃
∆𝑇 =?
𝑄 = 𝑚𝑐∆𝑇
∆𝑇 =
𝑄
35,000 𝐽
=
= 169 ℃
𝑚𝑐 (538.0 𝑔)(0.385 𝐽⁄𝑔 ∙ ℃)
5.) The temperature of a 55 gram sample of a certain metal drops by 113 oC as it loses 3500
Joules of heat. What is the specific heat of the metal?
𝑚 = 55 𝑔
∆𝑇 = −113 ℃
𝑄 = −3,500 𝐽
𝑐 =?
𝑄 = 𝑚𝑐∆𝑇
𝑐=
𝑄
−3,500 𝐽
=
= 0.56 𝐽⁄𝑔 ∙ ℃
𝑚∆𝑇 (55 𝑔)(−113 ℃)