An object with a mass of 40 g is dropped into 660 mL of water at 0^@C. If the object cools by 72 ^@C and the water warms by 3 ^@C, what is the specific heat of the material that the object is made of?

Feb 9, 2016

2.88 KJ/Kg.K

Explanation:

We assume there is no energy exchange with the surroundings and that the water and object is in thermal equilibrium and there is no phase change.

The heat energy absorbed by the water must equal that lost by the object when reaching thermal equilibrium.

There is one more piece of information we need to solve this problem and this is the heat capacity of water which is
4.184 KJ/Kg.K

The heat capacity basically indicates how much energy is needed to raise 1 Kg of water by 1 Kelvin. The equation that is used is
$Q = m {c}_{p} \delta T$

The water temperature changes from 0 to 3 degrees C. Which is the same as 3 K.

If we take the density of water to be 1000 g/ml then the heat absorbed is

Q= (0.66 Kg)(4.184 KJ/Kg.K)(3 K)
= 8.28 KJ

We use the same equation for the object but now solving for the heat capacity and taking the heat lost to be 8.28 KJ. The objects mass is 0.040 Kg and the temperature change is 72 K.

Hence the objects specific heat capacity is 2.88 KJ/Kg.K