Question

A calorimeter has 45.5 g of water with a temperature of 23.0°c. A separate amount of water with a mass of 53.5 g was heated to 87.0°c and poured into the calorimeter. The final temperature of the mixture is 57.0°c. Find the heat capacity of the calorimeter (in J/°C )

Answer 1

To find the heat capacity of the calorimeter, we can use the principle of heat transfer. The heat capacity of the calorimeter is 3.317 J/°C.

Step-by-step explanation:

To find the heat capacity of the calorimeter, we can use the principle of heat transfer, which states that the heat lost by the hot water is equal to the heat gained by the cold water and calorimeter.

First, we calculate the heat lost by the hot water using the equation Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Then, we equate this to the heat gained by the cold water and calorimeter, and solve for the heat capacity of the calorimeter.

Using the given values, the heat lost by the hot water is (53.5 g)(4.18 J/g°C)(87.0°C - 57.0°C) = 6893.1 J.

Since the heat lost by the hot water is equal to the heat gained by the cold water and calorimeter,

we have 6893.1 J = (45.5 g + 53.5 g + Ccal)(4.18 J/g°C)(57.0°C - 23.0°C), where Ccal is the heat capacity of the calorimeter.

Simplifying the equation, we get 6893.1 J = 2602.6 J + (4.18 J/g°C)(99.0°C - 23.0°C)(Ccal),

which gives us Ccal = 3.317 J/°C.

Therefore, the heat capacity of the calorimeter is 3.317 J/°C.