Help with the answer please

Question

A 66.63 g sample of iron (with a heat capacity of 0.450 J/g·°C) is heated to 100.0 °C. It is then transferred to a coffee cup calorimeter containing 38.74 g of water (specific heat of 4.184 J/g·°C) initially at 20.63 °C. If the final temperature of the system is 23.59 °C, how much heat was absorbed by the calorimeter?

Record your answer as a whole number in joules, and assume the sign is positive.

Answer 1

Heat lost by the hot iron MUST equal heat gained by the water and the calorimeter.

First we'll find the heat lost by the hot iron:

q = mC∆T = (66.63 g)(0.450 J/gº)(100.0 - 23.59) = 2291 J

Next, we'll find the heat gained by the water in the calorimeter:

q = mC∆T = (38.74 g)(4.184 J/gº)(23.59º - 20.63º) = 480 J

Assuming no heat is lost to the surroundings, the difference is the heat gained by the calorimeter.

2291 J - 480 J = 1811 J absorbed by the calorimeter