# An object with a mass of 30 g is dropped into 750 mL of water. If the object cools by 15 ^@C and the water warms by 15 ^@C, what is the specific heat of the material that the object is made of?

Mar 1, 2017

The specific heat is $= 104.65 k J k {g}^{-} 1 {K}^{-} 1$

#### Explanation:

The heat transferred from the hot object, is equal to the heat absorbed by the cold water.

For the cold water,  Delta T_w=15º

For the object DeltaT_o=15º

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

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

${m}_{0} {C}_{o} \cdot 15 = {m}_{w} \cdot 4.186 \cdot 15$

$0.030 \cdot {C}_{o} \cdot 15 = 0.75 \cdot 4.186 \cdot 15$

${C}_{o} = \frac{0.75 \cdot 4.186 \cdot 15}{0.030 \cdot 15}$

$= 104.65 k J k {g}^{-} 1 {K}^{-} 1$