Question

A 309 g block of copper at a temperature of 73.0°C is dropped into 579 g of water at 26.9°C. The water is contained in a 118 g glass container. Assume the glass has the same initial temperature as the water (26.9°C). What is the final temperature of the mixture?

The specific heat of copper is 387 J/kg˚C, the specific heat of water is 4186 J/kg˚C, and the specific heat of glass is 837 J/kg˚C.

Answer 1

given
m_Copper = 309 g = 0.309 kg
m_water = 579 g = 0.579 kg
m_glass = 118 g = 0.118 kg

C_copper = 387 J/(kg K)
C_water = 4186 J/(kg K)
C_glass = 837 J/(kg J)

let T is the final temperature of the mixture

m_copper*C_copper*(73 - T) = m_water*C_water*(T - 26.9) + m_glass*C_glass*(T - 26.9)

0.309*387*(73 - T) = 0.579*4186*(T - 26.9) + 0.118*837*(T - 26.9)

==> T = 29.0 degrees Celcius