Question

When 66.0 g of an unknown metal at 28.5 °C is placed in 83.0 g H2O at 78.5 °C, the water temperature decreases to 75.9 °C. What is the specific heat capacity of the metal? The specific heat capacity of water is 4.184 J/g⋅K.

Answer 1

Solution.

$Q(H2O) = c(H2O) \times m(H2O) \times (T2-T1)$

$Q(metal) = c(metal) \times m(metal) \times (T2-T3)$

$Q(H2O) = Q(metal)$

$Q(H2O) = 4.184 \times 83.0 \times (351.5 - 348.9) = -902.9 \ J$

$C(metal) = -\frac{Q(H2O)}{m(metal) \times (T2-T3)}$

$C(metal) = -\frac{902.9}{66.0 \times (348.9-301.5)} = 0.2886 \ \frac{J}{g \times K}$

Answer:

$C(metal) = 0.2886 \ \frac{J}{g \times K}$