Question

The temperature of air in a foundry increases when molten metals cool and solidify. Suppose 9.9 × 106 J of energy is added to the surrounding air by the solidifying metal. The air’s temperature increases by 55 K, and the air has a specific heat capacity of 1.0 × 103 J/kg • K. What is the mass of the heated air?

Answer 1

To find the mass of the heated air, you can use the formula that relates heat added, mass, specific heat capacity, and temperature change:

\[
Q = mc\Delta T
\]

where:
- \( Q \) = heat added (in joules)
- \( m \) = mass of the substance (in kg)
- \( c \) = specific heat capacity (in J/kg·K)
- \( \Delta T \) = change in temperature (in K)

You are given:
- \( Q = 9.9 \times 10^6 \, \text{J} \)
- \( c = 1.0 \times 10^3 \, \text{J/kg·K} \)
- \( \Delta T = 55 \, \text{K} \)

We need to rearrange the formula to solve for \( m \):

\[
m = \frac{Q}{c \Delta T}
\]

Now plug in the values:

\[
m = \frac{9.9 \times 10^6 \, \text{J}}{(1.0 \times 10^3 \, \text{J/kg·K})(55 \, \text{K})}
\]

Calculating the denominator:

\[
c \Delta T = (1.0 \times 10^3 \, \text{J/kg·K})(55 \, \text{K}) = 55 \times 10^3 \, \text{J/kg}
\]

Now substituting this back into the equation for \( m \):

\[
m = \frac{9.9 \times 10^6 \, \text{J}}{55 \times 10^3 \, \text{J/kg}} = \frac{9.9 \times 10^6}{55 \times 10^3}
\]

Calculating this gives:

\[
m = \frac{9.9}{55} \times 10^{6-3} = \frac{9.9}{55} \times 10^3 \, \text{kg}
\]

Calculating \( \frac{9.9}{55} \):

\[
\frac{9.9}{55} \approx 0.18
\]

So:

\[
m \approx 0.18 \times 10^3 \, \text{kg} = 180 \, \text{kg}
\]

Thus, the mass of the heated air is approximately **180 kg**.