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Question
A burner raises the temperature of 360 g of water from 40°C to 100°C in 5 minutes. Calculate the rate of heat supplied by the burner.
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Solution
m = 360 g = O .36 kg
Change in temperature, ΔT = (100 - 40)°C = 60°C = 60 K
Amount of heat required, Q = m x C x ΔT
= 0.36 x 4200 x 60 = 90720 J
Time taken = 5 min = 300 sec
Rate of heat supplied = `"Q"/"t" = 90720/300` = 302.4 J/s
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