Question

Question: Calculate the heat capacity, in joules and in calories per degree, of the following:

(a) 28.4 g of water

(b) 1.00 oz of lead

Short answer

The heat capacity in joules:

1. 28.4 g of water = 118.2$\frac{J}{{}^{0}C}$
2. 1.00 oz of lead =3.68 $\frac{J}{{}^{0}C}$

The heat capacity in calories:

1. 28.4 g of water =28.38 $\frac{cal}{{}^{0}C}$
2. 1.00 oz of lead =0.879 $\frac{cal}{{}^{0}C}$

Step-by-step solution

Heat capacity

The heat capacity of a substance can be calculated by the formula H = mc, where “m” is the mass of the substance and “c” is its specific heat capacity.

Heat capacity of water in joules

From table 5.1, we know that the specific heat capacity of water is 4.184J/g℃.

Mass of the water = 28.4 g (given).

Therefore, the heat capacity of 28.4 g of water in joules = 28.4 $\times$4.184

= 118.82$\frac{J}{{}^{0}C}$

Heat capacity of water in calories per degree

1 calorie = 4.186 J.

The heat capacity of 28.4 g of water in calories is evaluated as:

=$\frac{118.82}{4.186}\frac{cal}{{}^{0}C}$

=28.38 $\frac{cal}{{}^{0}C}$

Step 4:Heat capacity of lead in joules

From table 5.1, we know that the specific heat capacity of lead is 0.130J/g℃.

The mass of the lead = 1 oz (given)

The mass of the lead in grams = 28.35$\times$1 = 28.35 g.

Therefore, the heat capacity of 28.35 g of lead in joules = 0.130$\times$28.35

= 3.68 $\frac{J}{{}^{0}C}$

Heat capacity of lead in calories per degree

1 calorie = 4.186 J.

The heat capacity of 28.35 g of lead in calories is calculated as:

$\begin{array}{c}=\frac{3.68}{4.186}\\ =0.879\frac{cal}{{}^{0}C}\end{array}$