# An object with a mass of 120 g is dropped into 720 mL of water at 0^@C. If the object cools by 24 ^@C and the water warms by 3 ^@C, what is the specific heat of the material that the object is made of?

##### 1 Answer
Sep 19, 2016

Specific heat of object is $0.75$.

#### 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 objece is $120 \times s \times 24$

and heat gained by water is $720 \times 1 \times 3$.

Now heat lost should be equal to heat gained

$120 \times s \times 24 = 720 \times 1 \times 3$

i.e. $s = \frac{720 \times 3}{120 \times 24} = \frac{6 \cancel{720} \times \cancel{3}}{\cancel{120} \times 8 \cancel{24}} = \frac{6}{8} = \frac{3}{4} = 0.75$

Hence specific heat of object is $0.75$.