hot water of mass 1o kg at 90 degree centrigrade is cooled for taking bath by mixing 20 kg of water at 20 degree centigrade what is the final temperature of water?specific heat capacity of water is 422j/kg degree centigrade
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Let the final temperature be 'T' and the specific heat capacity of water be 'c'.
Heat lost = Heat gained
10 * c * (90 - T) = 20 * c * (T - 20)
T = 43.333
Heat lost = Heat gained
10 * c * (90 - T) = 20 * c * (T - 20)
T = 43.333
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