A block of silicon weighing 24.9 g is heated to 64.7 °C in boiling methanol

Question

A block of silicon weighing 24.9 g is heated to 64.7 °C in boiling methanol. It is then dropped into 28.2 g of methanol initially at 28.1 °C. Find the final temperature of silicon + methanol in °C.

Answer 1

q = mC∆T

q = heat

m = mass

C = specific heat

∆T = change in temperature

heat lost by hot silicon = heat gained by cool methanol

Looking up specific heat of silicon and methanol, I found C_silicon = 0.71 J/gº and C_methanol = 2.53 J/gº)

heat lost by hot silicon = q = (24.9 g)(0.71 Jº)(64.7 - Tf)

heat gained by cool methanol = q = (28.2 g)(2.53 J/gº)(Tf - 28.1)

(24.9 g)(0.71 Jº)(64.7 - Tf) = (28.2 g)(2.53 J/gº)(Tf - 28.1)

1144 - 17.7Tf = 71.3Tf - 2005

89Tf = 3149

Tf = 35.4ºC