What should be the temperature of 1g of ice, so that when it’s added to 9g of water, lowers the temperature of water by 10°C? Sw = 1cal/ g°C, Sice = 0.5cal/ g°C, Lfus= 80cal/g. (a) 20°C (b) -20°C (c) 0°C (d) 10°C
Correct answer is (b) -20°C For explanation: Let the initial temperature of ice be T°C. Heat lost by ice will be the same as the heat lost by water. ∴ 1*0.5*(0-T) + 1*80 = 9*1*10 ⇒ T = -20°C