# An object with a mass of 21 g is dropped into 400 mL of water at 0^@C. If the object cools by 150 ^@C and the water warms by 14 ^@C, what is the specific heat of the material that the object is made of?

Feb 19, 2016

The heat energy lost by the object is the same as the energy gained by the water, or ${Q}_{o} = {Q}_{w}$

#### Explanation:

${c}_{o} \cdot {m}_{o} \cdot \Delta {T}_{o} = {c}_{w} \cdot {m}_{w} \cdot \Delta {T}_{w}$

Now fill in what we know (${c}_{w} = 4.2$ and $1 m L = 1 g$ water):
${c}_{o} \cdot 21 \cdot 150 = 4.2 \cdot 400 \cdot 14$

Work this out:
${c}_{o} = \frac{4.2 \cdot 400 \cdot 14}{21 \cdot 150} \approx 7.47$

This is a ridiculously high outcome, as no known material has a specific heat this high. Probably a $0$ was lost or added in the problem.
With a mass of $210 g$ or an amount of water of $40 m L$ the outcome would be $0.747$, which is more plausible.