# An object with a mass of 120 g is dropped into 810 mL of water at 0^@C. If the object cools by 48 ^@C and the water warms by 8 ^@C, what is the specific heat of the material that the object is made of?

Dec 31, 2017

The specific heat is $= 4.71 k J k {g}^{-} 1 {K}^{-} 1$

#### Explanation:

The heat is transferred from the hot object to the cold water.

For the cold water,  Delta T_w=8ºC

For the object DeltaT_o=48ºC

${m}_{o} {C}_{o} \left(\Delta {T}_{o}\right) = {m}_{w} {C}_{w} \left(\Delta {T}_{w}\right)$

The specific heat of water ${C}_{w} = 4.186 k J k {g}^{-} 1 {K}^{-} 1$

Let, ${C}_{o}$ be the specific heat of the object

The mass of the object is ${m}_{o} = 0.12 k g$

The mass of the water is ${m}_{w} = 0.81 k g$

$0.12 \cdot {C}_{o} \cdot 48 = 0.81 \cdot 4.186 \cdot 8$

${C}_{o} = \frac{0.81 \cdot 4.186 \cdot 8}{0.12 \cdot 48}$

$= 4.71 k J k {g}^{-} 1 {K}^{-} 1$