# An object with a mass of 60 g is dropped into 900 mL of water at 0^@C. If the object cools by 40 ^@C and the water warms by 16 ^@C, what is the specific heat of the material that the object is made of?

##### 1 Answer
Mar 6, 2016

Specific heat of the material that the object is $6$.

#### Explanation:

In such cases heat lost by one equals heat gained by other. Heat lost / gained is given by product of mass $m$, specific heat $s$ (which is $1$ for water) and change in temperature,

Let the specific heat of the mass of $60 g$, which cools by ${40}^{o} C$ be $s$. Hence heat lost is $60 \times 40 \times s$ calories.

900mL of water means $900 g$ and it warms by 16∘C. Hence heat gained by it is $900 \times 16 = 14400$ calories.

Hence $60 \times 40 \times s = 14400$ or $x = \frac{14400}{60 \times 40} = 6$

Hence, specific heat of the material that the object is $6$.