Question

A 40 g block of ice is cooled to -78°C. and is then added to 610 g of water in an 80 g copper calorimeter at a temperature of 26°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g ·°C = 2090 J/kg°C

Answer 1

mi = 40g
Ti = -78 C
mw = 610 g
mc = 80 g
T = 26 C

Heat needed to completely melt the ice,
Q = mi*Ci*78 + mi*Lf
Q = 40 * 2.108 * 78 + 40 * 333.55 J
Q = 19919 J

Heat given by water & Calorimeter,
Q = mw*Cw*T + mc * Cc * T
Q = 610 * 4.186 * 26 + 80 * 0.386 * 26 J
Q = 67192.84 J

As the heat provided by water & copper calorimeter > Heat neede to melt the ice , so ice will completely melt !!
Let the final Temp be Tf,
mw*Cw*(T -Tf) + mc * Cc * (T-Tf) = 19919 + mi*Cw*Tf
610 * 4.186 * (26 - Tf) + 80 * 0.386 * (26 - Tf) = 19919 + 40 * 4.186 * Tf
Tf = 17.2^o C