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प्रश्न
A liquid of mass 0.2 kg and temperature 135°C is cooled to 25°C. If the specific heat capacity of the liquid is 750 J kg−1°C−1, find the heat energy given out. [Hint: θF = (135 – 25) = 110°C]
संख्यात्मक
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उत्तर
Fall in temperature θF = (135− 25) = 110°C
∴ Heat = mcθF
= 0.2 kg × 750 J kg−1°C−1 × 110°C
= 16500 J
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