# 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?

Jan 29, 2018

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:

$\sum Q = 0$,

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

Since $Q = m c \Delta T$, then:

${Q}_{o b j e c t} + {Q}_{w a t e r} = 0$;

${m}_{o b j e c t} {c}_{o b j e c t} \Delta {T}_{o b j e c t} + {m}_{w a t e r} {c}_{w a t e r} \Delta {T}_{w a t e r} = 0$.

Then:

$120 g \cdot {c}_{o b j e c t} \cdot \left(- {48}^{o} C\right) + 640 g \cdot 1 \frac{c a l}{{g}^{o} C} \cdot {3}^{o} C = 0$,

where I have used the density of water (1 g/ml) to estimate its mass, and ${c}_{w a t e r} = 1 \frac{c a l}{{g}^{o} C} .$

Finding the value for ${c}_{o b j e c t} :$

${c}_{o b j e c t} = \frac{1920}{5760} \frac{c a l}{{g}^{o} C}$;

${c}_{o b j e c t} = 0.33 \frac{c a l}{{g}^{o} C}$, approximately.