A calorimeter of heat capacity 83.72JK^(-1) contains 0.48 kg of water at 35^(@)C. How much mass of ice at 0^(@)C should be added to decrease the temperature of the calorimeter to 20^(@)C.

A calorimeter of heat capacity 83.72JK^(-1) contains 0.48 kg of water

1.	A calorimeter of heat capacity 83.72JK^(-1) contains 0.48 kg of water at 35^(@)C. How much mass of ice at 0^(@)C should be added to decrease the temperature of the calorimeter to 20^(@)C.
Answer» <html><body><p></p>Solution :Let .m. be the mass of <a href="https://interviewquestions.tuteehub.com/tag/ice-1035302" style="font-weight:bold;" target="_blank" title="Click to know more about ICE">ICE</a> added, <br/> Heat gained by ice during melting = <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> <br/> = `mxx0.335xx10^(6)<a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a>` <br/> Heat gainerd by molten ice (water) during rise in temperature = `mxx4186xx(20-0)=mxx83720J` <br/> `therefore` Heat gained by the ice = Heat <a href="https://interviewquestions.tuteehub.com/tag/lost-537630" style="font-weight:bold;" target="_blank" title="Click to know more about LOST">LOST</a> by the <a href="https://interviewquestions.tuteehub.com/tag/calorimeter-908024" style="font-weight:bold;" target="_blank" title="Click to know more about CALORIMETER">CALORIMETER</a> and water <br/> `mxx[0.335xx10^(6)+83720]=1255.8+30139.2=31395J` <br/> `m[335000+83720]=31395` <br/> `m=31395/418720=0.07498kg`.</body></html>