# An object with a mass of 160 g is dropped into 880 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 12 ^@C, what is the specific heat of the material that the object is made of?

Feb 24, 2017

The specific heat is $= 2.30 k J k {g}^{-} 1 {K}^{-} 1$

#### 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=12º

For the object DeltaT_o=120º

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

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

${m}_{0} {C}_{o} \cdot 120 = {m}_{w} \cdot 4.186 \cdot 12$

$0.160 \cdot {C}_{o} \cdot 120 = 0.88 \cdot 4.186 \cdot 12$

${C}_{o} = \frac{0.88 \cdot 4.186 \cdot 12}{0.160 \cdot 120}$

$= 2.30 k J k {g}^{-} 1 {K}^{-} 1$