N1 Engineering Science|The Easy Way! 75 Example 4.6
Determine the specific heat capacity of a certain piece ofmetal with amass of 20 kg. It was found by experiment that after the piece ofmetal received 1,85 MJ of heat energy, its temperature rose by 230 °C.
Solution: c = ?
m= 20 kg
Q = 1,85 × 106 MJ 3t = 230 °C
Q =mc3t
1,85 × 106 = 20 × c × 230 1,85 × 106 = 4 600 × c
\ c = 185106 , ×
4 600 = 402,17 J/kg °C 4.8 Heat transfer (substances in contact)
When a hot substance is mixed or brought into contact with a cooler one, heat is transferred from the hotter substance to the cooler one until their temperatures are equal.
Assuming that no heat is lost during the transfer: Heat lost (hot body) = Heat gained (cooler body) ................................................... (4.4)
or mc3t (hot body) = mc3t (cold body) ................................................................. (4.5) Example 4.7
After a steel component was welded, its temperature had risen to 850 °C. It was quenched in a tank containing 30 ℓ of water, with an initial temperature of 19,5 °C, until both the component and the water reached an equal temperature of 25 °C.
Assuming that no heat was lost and that the specific heat capacity of the steel and the water is 494 J/kg °C and 4 200 J/kg °C respectively, calculate themass of the steel component.
Solution: Steel
t1s = 850 °C t2 = 25 °C
c = 494 J/kg °C
Water
m= 30 kg (1 ℓ = 1 kg) t1w = 19,5 °C
t2 = 25 °C c = 4 200 J/kg °C
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