550g of ice initially at 100C are dropped into 325g of water

55.0g of ice initially at -10.0°C are dropped into 325g of water initially at 45.0 °C in a thermally isolated container. What is the final trmperature when thermal equilibrium is reached?

Solution

Amount of heat the ice can absorb while getting heated upto 0 degree celsius would be given as the product of the specific heat capacity, mass and temperature change.

hence, the heat the ice will absorb while getting heated to 0 C = 55 x 10 x 2..03 = 1116.5 J

This will bring the temperature of water down by: t = 1116.5 / 325 x 4.179 = 0.8221 degree celsius.

Now, the ice can absorb heat for it to melt which is given as 333.55 x 55 = 18345.25 J

This will again cause water to cool down by: 18345.25 / 325 x 4.179 = 13.5073

So the net drop in temperature for water = 14.3294 degree centigrade.

Hence the temperature of water by the time, ice melts would be 30.6706

Now, we will have 55 g of water a 0 degree celsius and 325 g of water at 30.6706 degree celsius.At equilibrium, the net heat lost would be equal to net heat gained.

hence, 55 x 4.179 (T - 0) = 325 x 4.179 x (30.6706 - T)

or, 55T = 9967.945 - 325T

or T = 26.2314 degree celsius.

Therefore the final temperature of the system is 26.2314 degree celsius