74 Chapter 4 • Heat


Te amount of heat energy a substance gains or loses is proportional to: • themass of the substance, • the specific heat capacity of the substance (i.e. the type ofmaterial), and • the change in temperature of the substance.


Tus Q = mc3t .............................................................................................................. (4.3) where Q = heat energy in joules (J) m = mass of the substance in kg c = specific heat capacity in J/kg °C


3t = change in temperature in °C or K (tf – ti) or (t2 – t1) t1 = initial temperature (or ti) t2 = final temperature (or tf)


Example 4.4


58,5 kJ of heat energy is absorbed by a copper cylinder with amass of 2 kg. The initial temperature was 20 °C and the specific heat capacity of copper is 390 J/kg °C. Calculate; (a) (b)


the rise in temperature, and the final temperature.


Solution: Q = 58 500 J m= 2 kg (a) \ 3t = Q


Q = mc3t mc


= 58 500 = 75 °C


Example 4.5


Calculate the amount of heat energy required to raise the temperature of 5 kg steel from–20 °C to 140 °C if the specific heat capacity of the steel is 0,486 kJ/kg °C.


Solution: Q = ?


Q = mc3t = 5 × 486 × 160 = 388 800 J = 388,8 kJ


2 390× (b) t1 = 20 °C 3t = t2 – t1


\ t2 = 3t + t1 = 75 + 20


= 95 °C c = 390 J/kg °C


m= 5 kg 3t = 140 – (–20) = 160° c = 486 J/kg °C


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