Question

Two 20.0-g ice cubes at –20.0 °C are placed into 255 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts.

Answer 1

Solution :-

First ice will change from -20 C to 0 C then it melts at 0 C and then liquid water will raise in temperature from 0 C so we can make the following set up

-q hot water = q ice + delta H fus + q cold water

-(255 g * 4.184 J per g C * (T2- 25.0 C)) = (40 g * 2.02 J per g C * 20 C) + (40 g * 334 J per )+(40 g * 4.184 J per g C * (T2-0 C))

-1066.92 T2 +26673 = 1616 + 13360 + 167.36 T2

26673 – 1616 – 13360 = 167.36 T2 + 1066.92 T2

11697 = 1234.28 T2

11697 / 1234.28 = T2

9.48 C= T2

So the final temperature is 9.48 C