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A large room in a house holds 947 kg of dry air at 33.2°C. A woman opens a window briefly and a cool breeze brings in an additional 62.4 kg of dry air at 13.2°C. At what temperature (in degrees Celsius) will the two air masses come into thermal equilibrium, assuming they form a closed system? (The specific heat of dry air is 1006 J/ kg * °C, although that value will cancel out of the calorimetry equation.)

°C

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boffeemadrid

Answer:

31.96362 °C

Explanation:

[tex]m_1[/tex] = Mass of air in the room = 947 kg

[tex]m_2[/tex] = Mass of air entering the room = 62.4 kg

[tex]T_1[/tex] = Temperature in the room = 33.2°C

[tex]T_2[/tex] = Temperature air entering the room = 13.2°C

T = Equilibrium temperature

c = Specific heat of air = 1006 J/kg °C

In the case of thermal equilibrium we have the relation

[tex]m_1c(T_1-T)=m_2c(T-T_2)\\\Rightarrow T=\frac{m_2T_2+m_1T_1}{m_1+m_2}\\\Rightarrow T=\frac{62.4(13.2+273.15)+947(33.2+273.15)}{947+62.4}\\\Rightarrow T=305.11362\ K=305.11362-273.15=31.96362^{\circ}C[/tex]

The temperature of thermal equilibrium is 31.96362 °C

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