A 0.9 kg block of ice is initially at a temperature of -5°C.

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Apr 16th, 2015

the heat capacity of ice is  2.108 kJ/ (kg*K) the latent heat of ice is 334.7 kJ/kg,

let the final temperature of the water is T, 

then 390kJ - 0.9*2.108 *5-0.9*334.7 =  79.3 kJ 

T = 79.3KJ/(0.9*4.184) = 21 C


Apr 16th, 2015

can u answer part b?

Apr 16th, 2015

yes, it need  to increase by a factor 3.6 too

Apr 16th, 2015

thanks :)

Apr 16th, 2015

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