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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 wateris 1 cal g^(-1@)C^(-1)) |
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Answer» Solution :Given mass of water =250 g Change intemperature `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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