Question:

A solid cylinder of aluminum with a mass of 2 kg and a specific heat capacity of 900 J/kg°C is initially at a temperature of 80°C. The cylinder is placed in 2 kg of water at 20°C. If the final temperature of the system is 30°C, calculate the initial temperature of the water. Assume no heat is lost to the surroundings.

Answer:

To solve this problem, we will use the principle of conservation of energy to calculate the initial temperature of the water.

First, let's calculate the heat transferred from the aluminum cylinder to the water using the equation:

q = mcΔT

Where: q = heat transferred m = mass c = specific heat capacity ΔT = change in temperature

Since no heat is lost to the surroundings, the heat lost by the aluminum cylinder is equal to the heat gained by the water.

For the aluminum cylinder:

q_aluminum = m_aluminum * c_aluminum * (T_initial - T_final)

Substitute the given values: q_aluminum = 2 kg * 900 J/kg°C * (T_initial - 30°C)

And for the water:

q_water = m_water * c_water * (T_final - T_initial)

Substitute the given values: q_water = 2 kg * 4186 J/kg°C * (30°C - T_initial)

Since q_aluminum = q_water, we can set the two equations equal to each other and solve for the initial temperature of the water:

2 kg * 900 J/kg°C * (T_initial - 30°C) = 2 kg * 4186 J/kg°C * (30°C - T_initial)

Simplify and solve for T_initial:

1800(T_initial - 30) = 8372(30 - T_initial) 1800T_initial - 54000 = 837230 - 8372T_initial 1800T_initial + 8372T_initial = 837230 + 54000 10172T_initial = 251160 T_initial = 24.7°C

The initial temperature of the water is approximately 24.7°C.