Physics Question

Question

A 500 g sample of copper at 80.0°C is placed in 2.00 kg of water at 25.0°C. The water is in a closed styrofoam container which does not exchange any heat with the environment. What is the temperature of the mixture after the two materials reach thermal equilibrium?

Answer 1

The key equation here is that the heat into the water is the same as the heat out of the copper. Therefore, if we let dT denote the increase in temperature of the water, then the change in temperature of the copper will be 80-(25+dT) = 55-dT,

and the specific heats of water and copper are c_water= 4.186 J/g*C and c_copper = .385 J/ g*C, then the

m_copper * c_copper * (55-dT) = m_water * c_water * dT,

dT=(55 *C * m_copper*c_copper )/ (m_water * c_water + m_copper * c_copper)

= (63.2 kJ) / (8398 J/C + 193 J/C )

= 7.357 degrees C. Thus the mixture is at 25+7.357= 32.357 degrees C.