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Calculate the amount of heat energy required to increase the temperature of 250 g of water from `27^(@)C` to `67^(@)C` (Specific heat capacity of water is 1 cal `g^([email protected])C^(-1)`) |
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Answer» Given mass of water =250 g Change in temperature `Delta`=final temperature -intitial temperature =`67^(@)C-27^(@)C=40^(@)C=40^(@)C` specific heat capacity of water =1 cal `gal^(-1).^@C^(-1)` Heat energy supplied ,Q =`MsDelta t` Q=`250xx1xx40`=10000 cal =kCal |
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