How many gram of ice at –14°C is needed to cool 200 g of water from 25°C to 10°C. Take specific heat of ice as 0.5 cal+g+°C, the latent heat of ice as 80 cal+g and specific heat of water as 1 cal+g+°C.
Dear student,
200 gm x 15 deg = 3000 cal
14 deg C x M gm x 0.5 cal/gm = 7.M calories
80 x M = 80.M cal
10 x M = 10M cal
So we have 3000 = 7xM + 80xM + 10xM = 97xM
and M = 3000 / 97 = 30.93 gm
Regards