3.75 X105 calories of heat is given out by 5 kg of water at 100o C. Calculate the temperature of cooled water. SHC of water =1000 calories / kg o C.
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q= mC∆t
3.75×10^5= 5*1000*(100-t)
solve for t
t= 25°C
3.75×10^5= 5*1000*(100-t)
solve for t
t= 25°C
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