# An object with a mass of 50 g is dropped into 300 mL of water at 0^@C. If the object cools by 80 ^@C and the water warms by 8 ^@C, what is the specific heat of the material that the object is made of?

##### 1 Answer
Dec 7, 2017

${C}_{o} = 2508 J \textcolor{w h i t e}{l} k {g}^{- 1} \textcolor{w h i t e}{l} {K}^{- 1}$

#### Explanation:

$\Delta E = m c \Delta \theta$, where:

• $\Delta E$ = change in energy ($J$)
• $m$ = mass of the object ($k g$)
• $c$ = specific heat capacity of the object ($J$ $k {g}^{- 1}$ ${K}^{- 1}$)
• $\Delta \theta$ = change in temperature ($K$)

Assuming all heat is transferred we have:
${m}_{o} {C}_{o} \Delta {\theta}_{o} = {m}_{w} {c}_{w} \Delta {\theta}_{w}$

${m}_{o} = 50 g = 0.05 k g$
$\Delta {\theta}_{o} = {80}^{\circ} C = 80 K$
${\rho}_{w} \approx 1000 k g \textcolor{w h i t e}{l} {m}^{- 3}$
${V}_{w} = 300 m L = 0.3 L = 0.0003 {m}^{3}$
${m}_{w} = 1000 \cdot 0.0003 = 0.3 k g$
$\Delta {\theta}_{w} = {8}^{\circ} C = 8 K$
${c}_{w} \approx 4180 J \textcolor{w h i t e}{l} k {g}^{- 1} \textcolor{w h i t e}{l} {K}^{- 1}$

$0.05 {C}_{o} \cdot 80 = 0.3 \cdot 4180 \cdot 8$

${C}_{o} = \frac{0.3 \cdot 4180 \cdot 8}{0.05 \cdot 80} = 2508 J \textcolor{w h i t e}{l} k {g}^{- 1} \textcolor{w h i t e}{l} {K}^{- 1}$