Question

You are using a calorimeter to calculate the specific heat capacity of a metallic ore. The calorimeter contains 0.50 kilograms of water at room temperature (22 °C). We heat the ore in boiling water and then drop the metal into the calorimeter and wait for the water and metal to reach the same temperature. The mass of the ore is 3.5 kilograms. We find that the water has increased in temperature to 24.3 °C. Recall that the specific heat of water is 4.18 J/g–°C. Calculate the specific heat capacity of the ore.

Answer 1

the specific heat capacity of the ore : 0.018 J/g°C  

Further explanation

In the calorimeter, the heat received is the same as the heat released  

Q abs = Q release

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Q released by a metallic ore and absorbed by water at calorimeter

Q ore = Q water

• Q water

m = 0.5 kg = 500 g

c = 4.18 J/g–°C.

Δt = 24.3 - 22 = 2.3

$\tt Q=m.c.\Delta t\\\\Q=500\times 4.18\times 2.3=4807~J$

• the specific heat capacity of the ore.

Q ore = Q water = 4807 J

m ore = 3.5 kg = 3500 g

Δt = 100 - 24.3 = 75.7

$\tt 4807=3500\times c\times 75.7\\\\c=0.018~J/g^oC$