# An object with a mass of 100 g is dropped into 500 mL of water at 0^@C. If the object cools by 50 ^@C and the water warms by 4 ^@C, what is the specific heat of the material that the object is made of?

Mar 4, 2017

The specific heat is $= 1.67 k J k {g}^{-} 1 {K}^{-} 1$

#### 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=4º

For the metal DeltaT_o=50º

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

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

${m}_{0} {C}_{o} \cdot 50 = {m}_{w} \cdot 4.186 \cdot 4$

$0.1 \cdot {C}_{o} \cdot 50 = 0.5 \cdot 4.186 \cdot 4$

${C}_{o} = \frac{0.5 \cdot 4.186 \cdot 4}{0.1 \cdot 50}$

$= 1.67 k J k {g}^{-} 1 {K}^{-} 1$