Question

The sample of metal is heated to 100°C before it is added to the calorimeter. The calorimeter is allowed to sit for five minutes, and the temperature is monitored until the overall system (both water and metal) reaches thermal equilibrium at a temperature of 15.5°C. You are given an 80-gram sample of metal and a small calorimeter determine the specific heat of the sample. Before adding the sample, 150 grams of water are added to the calorimeter, and the starting temperature of the water is measured to be 14.1°C. You know that the specific heat of water is approximately 4.184 J/g C. Use the heat absorbed by the water to determine the specific heat of the metal.

Answer 1

To determine the specific heat of the metal, the heat absorbed by the water is calculated and then assumed to equal the heat lost by the metal. This value is used to compute the metal's specific heat, resulting in 0.130 J/g°C.

Step-by-step explanation:

To determine the specific heat of the metal sample using the calorimeter, we apply the principle of heat exchange, where the heat lost by the metal is equal to the heat gained by the water. The specific heat of water is 4.184 J/g°C.

First, calculate the heat absorbed by the water using the equation Q = mcΔT, where Q is the heat absorbed, m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature.

Qwater = mwater × cwater × ΔTwater

Qwater = 150 g × 4.184 J/g°C × (15.5°C - 14.1°C)

Qwater = 150 g × 4.184 J/g°C × 1.4°C

Qwater = 879.36 J

Next, assume all the heat lost by the metal has been gained by the water:

Qmetal = -Qwater

Then calculate the specific heat of the metal using the heat lost by the metal:

cmetal = -Qwater / (mmetal × ΔTmetal)

ΔTmetal = Tinitial,metal - Tfinal,metal

ΔTmetal = 100°C - 15.5°C

ΔTmetal = 84.5°C

cmetal = -879.36 J / (80 g × 84.5°C)

cmetal = -0.130 J/g°C

The negative sign indicates heat loss from the metal, hence the magnitude of 0.130 J/g°C gives us the specific heat of the metal.