Answer to Question #75005 in General Chemistry for Kevin

Question

A 3.50kg of block of copper at a temperature of 80°C is dropped into a 900g brass calorimeter containing 1.20kg of water at 21°C.
A. What is the final temperature of the system?
B. How much heat is lost or gained by the system?

Answer 1

Q = mc∆T

Specific heat of water 4186 J/kg·°C, specific heat of copper 387 J/kg·°C.

Qhot = mCu·cCu·(Tf – 80°C)

Qcold = mw·cw·(Tf – 21°C)

Qhot + Qcold = 0

Qcold = - Qhot

mw·cw·(Tf – 21°C) = - mCu·cCu·(Tf – 80°C)

Tf(mwcw + mCucCu) = mCucCu(80°C) + mwcw(21°C)

Tf = (mCucCu(80°C) + mwcw(21°C)) / (mwcw + mCucCu) = (4.4kg·387J/kg·°C·80°C + 1.2kg·4186J/kg·°C·21°C) / (4.4kg·387J/kg·°C + 1.2kg·4186J/kg·°C) = 35.94 °C

QCu = 4.4kg·387J/kg·C·44.06°C = 75025.4J = 75 kJ

Qw = 1.2kg·4186J/kg·°C·14.94°C = 75046.6J = 75 kJ