Question

A piece of copper weighing 400 g is heated to 100 oC and then quickly transferred to a copper calorimeter of mass 10 g containing 100 g of liquid of unknown specific heat capacity at 30 oC. If the final temperature of the mixture is 50 oC, calculate the specific heat capacity of the liquid. (Specific heat capacity of copper = 390 J kg-1 K-1)?

Answer 1

To calculate the specific heat capacity of the liquid, use the principle of conservation of energy. The heat lost by the copper piece is equal to the heat gained by the liquid and the calorimeter. Set up an equation and solve for the specific heat capacity of the liquid.

Step-by-step explanation:

To calculate the specific heat capacity of the liquid, we can use the principle of conservation of energy. The heat lost by the copper piece is equal to the heat gained by the liquid and the calorimeter. The heat lost by the copper piece can be calculated using the formula: Q = mcΔT, where Q is the heat lost in Joules, m is the mass of the copper piece in kg, c is the specific heat capacity of copper in J/kg * K, and ΔT is the change in temperature.

First, calculate the heat lost by the copper piece:

Q(copper) = mcΔT = (0.4 kg)(390 J/kg * K)(100 °C - 50 °C)

Next, calculate the heat gained by the liquid and the calorimeter:

Q(liquid + calorimeter) = (m(liquid) + m(calorimeter))c(liquid + calorimeter)ΔT = (0.1 kg + 0.1 kg)(c(liquid) + 390 J/kg * K)(50 °C - 30 °C)

Since the heat lost by the copper piece is equal to the heat gained by the liquid and the calorimeter, we can set up the equation:

Q(copper) = Q(liquid + calorimeter)

Solving for c(liquid), the specific heat capacity of the liquid, we get:

c(liquid) = (Q(copper) - Q(liquid + calorimeter))/(m(liquid)ΔT)

Substituting the known values, we get:

c(liquid) = [(0.4 kg)(390 J/kg * K)(100 °C - 50 °C) - (0.1 kg + 0.1 kg)(390 J/kg * K)(50 °C - 30 °C)]/(0.1 kg(50 °C - 30 °C))

Calculating this expression gives us the specific heat capacity of the liquid.