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.
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