If 335 g of water at 65.5°C loses 9750 J heat, what is the final temperature of the water?

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Answer 1 · 2017-03-26T07:38:02

The final temperature is #=58.6ºC#

Let the final temperature be #=TºC#

Then,

#DeltaT=(65.5-T)ºC#

Mass of water, #m=0.335kg#

Heat lost, #H=9.750kJ#

specific heat of water, #s=4.186kJkg^-1ºC^-1#

Therefore,

#H=msDeltaT#

#9.750=0.335*4.186*(65.5-T)#

#65.5-T=9.750/(0.335*4.186)=6.95#

#T=65.5-6.95=58.6ºC#