# An object with a mass of 125 g is dropped into 750 mL of water at 0^@C. If the object cools by 40 ^@C and the water warms by 3 ^@C, what is the specific heat of the material that the object is made of?

Apr 19, 2016

${c}_{o b j e c t} = 1.88 \frac{J}{g \setminus {\quad}^{o} C} = 0.45 \setminus \frac{\textrm{c a l}}{g \setminus {\quad}^{o} C}$

#### Explanation:

Because energy is conserved, the heat gained by the water equals the heat lost by the object.

${Q}_{o b j e c t} = - {Q}_{w a t e r}$

The minus sign is there because one object is losing heat while the other is absorbing heat.

The specific heat of water is known, $\setminus \quad {c}_{w a t e r} = 4.18 \frac{J}{g \setminus {\quad}^{o} C}$

1 mL of water is the same as 1 gram of water so ${m}_{w a t e r} = 750 g$

We have that $Q = m c \setminus \Delta T$, so

${Q}_{o b j e c t} = - {Q}_{w a t e r}$

${m}_{o b j e c t} \setminus \quad {c}_{o b j e c t} \setminus \quad \setminus \Delta {T}_{o b j e c t} = - {m}_{w a t e r} \setminus \quad {c}_{w a t e r} \setminus \quad \setminus \Delta {T}_{w a t e r}$

Substitute known values and obtain: $\setminus \quad {c}_{o b j e c t} = 1.88 \frac{J}{g \setminus {\quad}^{o} C}$

If you want the units of the answer to be in terms of calories, $\setminus \frac{\textrm{c a l}}{g \setminus {\quad}^{o} C}$, use the unit conversion $1 c a l = 4.18 J$

${c}_{o b j e c t} = 0.45 \setminus \frac{\textrm{c a l}}{g \setminus {\quad}^{o} C}$