Question

A 49-g block of copper at -11 ∘C
is added to 140 g of water in a 74-g aluminum cup. The cup and the water have an initial temperature of 4.1 ∘C. Find the equilibrium temperature of the cup and its contents.

Answer 1

given
m_copper = 49 g = 0.049 kg
m_water = 140 g = 0.140 kg
m_alu = 74 g = 0.074 kg

we know,
C_copper = 386 J/(kg K)
C_aluminum = 900 J/(kg K)
C_water = 4186 J(kg K)

let T is final equilibrium temperature

heat gained by copper = heat lost by water and aluminum

m_copper*C_copper*(T - (-11)) = m_water*C_water*(4.1 - T) + m_aluminum*C_aluminum*(4.1 - T)

0.049*386*(T + 11) = 0.14*4186*(4.1 - T) + 0.074*900*(4.1 - T)

==> T = 3.67 degrees Celsius <<<<<<<----------Answer