A 0.2 kg piece of unknown metal is heated to 87 °C and then dropped into a 0.28 kg sample of water at 22°C. If the final temperature of the mixture is 24°C, what is the specific heat of the unknown metal?

Mar 2, 2017

The specific heat of the metal is $= 0.186 k J k {g}^{-} 1 {K}^{-} 1$

Explanation:

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

For the cold water,  Delta T_w=24-22=2º

For the metal DeltaT_o=87-24=63º

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

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

${m}_{0} {C}_{o} \cdot 63 = {m}_{w} \cdot 4.186 \cdot 2$

$0.2 \cdot {C}_{o} \cdot 63 = 0.28 \cdot 4.186 \cdot 2$

${C}_{o} = \frac{0.28 \cdot 4.186 \cdot 2}{0.2 \cdot 63}$

$= 0.186 k J k {g}^{-} 1 {K}^{-} 1$