# An object with a mass of 32 g is dropped into 480 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 4 ^@C, what is the specific heat of the material that the object is made of?

Sep 23, 2016

Specific heat of object is $0.50$.

#### Explanation:

Let the specific heat of object be $s$.

As quantum of heal gained / lost is given by $m \times s \times \left({t}_{2} - {t}_{1}\right)$, where $m$ is mass of object gaining / loosing heat and $s$ is its specific heat and $\left({t}_{2} - {t}_{1}\right)$ is difference in temperature caused by heating / cooling,

heat lost by object is $32 \times s \times 120$

and heat gained by water is $480 \times 1 \times 4$.

Now heat lost should be equal to heat gained

$32 \times s \times 120 = 480 \times 1 \times 4$

i.e. $s = \frac{480 \times 4}{32 \times 120} = \frac{\cancel{4} \times \cancel{120} \times \cancel{4}}{\cancel{4} \times \cancel{4} \times 2 \times \cancel{120}} = \frac{1}{3} = 0.50$

Hence specific heat of object is $0.50$.