Question

An object with a mass of `120 g` is dropped into `640 mL` of water at `0^@C`. If the object cools by `48 ^@C` and the water warms by `3 ^@C`, what is the specific heat of the material that the object is made of?

Answer 1

0.33 cal/gºC

Explanation:

At equilibrium, the final temperature `T` of both the object and the mass of water must be the same. In addition, in this condition:

`sumQ = 0`,

where `Q` indicates the amount of heat that is absorbed/lost by the two bodies in the system.

Since `Q = mcDeltaT`, then:

`Q_(object) + Q_(water) = 0`;

`m_(object)c_(object)DeltaT_(object) + m_(water)c_(water)DeltaT_(water) = 0`.

Then:

`120 g * c_(object) * (-48^oC) + 640 g * 1 (cal)/(g^oC) * 3^oC = 0`,

where I have used the density of water (1 g/ml) to estimate its mass, and `c_(water) = 1 (cal)/(g^oC).`

Finding the value for `c_(object):`

`c_(object) = (1920)/(5760)(cal)/(g^oC)`;

`c_(object) = 0.33 (cal)/(g^oC)`, approximately.