Question

A 60.0 g aluminum block, initially at 55.00 °C, is submerged into an unknown mass of water at 293.15 K in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 25.00 °C. What is the approximate mass of the water? The specific heat of water is 4.18 J/g . °C. The specific heat of aluminum is 0.897 J/g . °C.

Answer 1

The approximate mass of the water is 80kg

Explanation: Heat lost=heat gained

M1c1(Ʃ)=M2c2(Ʃ)

M1 is mass of aluminum

M2 is the mass of water

C1 is specific heat capacity of aluminum

C2 is specific heat capacity of water

Ʃ is change in temperature.

60 x0.897 x(55-25)=M2 x 4.18 x (25-20.15)

1614.6=20.27M2

M2=79.65

M2=80kg