Assuming that the half and half mixture was at 20°C initially how much heat must
ID: 1778148 • Letter: A
Question
Assuming that the half and half mixture was at 20°C initially how much heat must be removed from it to cool it down to the temperature of the ice+salt mixture? Use the total mass of the mixture as 125 g and take its specific heat as 0.8 cal/(g °C) (The specific heat of cream is about 0.6 cal/(g°C) and that of milk is close to that of water, 1 cal/(g °C), so the average of approximately equal wt. mixture of the two gives 0.8 cal/(g °C)). Assuming that the latent heat of ice in a salt+water mixture is the same as that of pure ice (80 cal/g), how much heat does it take to melt 600 g of ice?
Explanation / Answer
Q= m C deltaT
Q = 125 x 0.8 x (20 - 0)
Q = 2000 cal ........Ans
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Q = m Lf
Q = 600 x 80
Q = 48000 cal .......Ans