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Question
50 g of metal piece at 27°C requires 2400 J of heat energy so as to attain a temperature of 327°C . Calculate the specific heat capacity of the metal.
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Solution
m = 50 g or `50/1000` kg
H = 2400 J
Q1 = 27 °C
Q2 = 327 °C
H = mcθ
c = `"H"/("m"theta)`
`= 2400/(50/1000 xx (327 - 27))`
`= 2400/(5/100 xx 300)`
`= 2400/15 = 160` J/Kg - K
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