# An object with a mass of 32 g is dropped into 480 mL of water at 0^@C. If the object cools by 80 ^@C and the water warms by 4 ^@C, what is the specific heat of the material that the object is made of?

May 18, 2017

The specific heat is $= 3.14 k J k {g}^{-} 1 {K}^{-} 1$

#### Explanation:

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

For the cold water,  Delta T_w=4ºC

For the object DeltaT_o=80ºC

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

${C}_{w} = 4.186 k J k {g}^{-} 1 {K}^{-} 1$

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

$0.032 \cdot {C}_{o} \cdot 80 = 0.48 \cdot 4.186 \cdot 4$

${C}_{o} = \frac{0.48 \cdot 4.186 \cdot 4}{0.032 \cdot 80}$

$= 3.14 k J k {g}^{-} 1 {K}^{-} 1$