Question

Two 20.0-g ice cubes at –20.0 °C are placed into 265 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

Two ice cubes = 2*20 = 40gm

first the 40gm of ice at -20C must warm to 0C(q= mcdelta T) then each must absorb 334j/K to melt (m*Hf) and finally the 40gm will heat to the same final temperature to which the 265gm of of 25C water cools (q = mcdeltaT)

40*20*2.027 + 40*334+ [40*(Tf-0C)*4.184]   = 265*(25-Tf)*4.184

1621.6 +13360 + 167.36Tf   = 27719-1108.76Tf

-12737.4+ 1276.13Tf = 0

-12737.4 = -1276.13Tf

Tf = 12737.4/1276.13 = 9.98C⁰

final temperature is 9.98C⁰