How long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘C? Note that the specific heat of air is 1006 J/(kg⋅∘C) and the density of air is 1.20kg/m3.

Express your answer numerically in minutes, to three significant figures.

Answer :

Answer : The time required is, 16.1 minutes.

Explanation :

First we have to calculate the amount of heat required to increase the temperature is:

[tex]Q=mC\Delta T\\\\Q=\rho VC\Delta T[/tex]

[tex](m=\rho V)[/tex]

where,

Q = amount of heat required = ?

m = mass

[tex]\rho[/tex] = density of air = [tex]1.20kg/m^3[/tex]

V = volume of air

C = specific heat of air = [tex]1006J/kg^oC[/tex]

[tex]\Delta T[/tex] = change in temperature = [tex]10.0^oC[/tex]

Now put all the given values in above formula, we get:

[tex]Q=\rho VC\Delta T[/tex]

[tex]Q=(1.20kg/m^3)\times (3.00m\times 5.00m\times 8.00m)\times (1006J/kg^oC)\times (10.0^oC)[/tex]

[tex]Q=1.449\times 10^6J[/tex]

Now we have to calculate the time required.

Formula used :

[tex]t=\frac{Q}{P}[/tex]

where,

t = time required = ?

Q = amount of heat required = [tex]1.449\times 10^6J[/tex]

P = power = 1500 W

Now put all the given values in above formula, we get:

[tex]t=\frac{1.449\times 10^6J}{1500W}[/tex]

[tex]t=966s\times \frac{1min}{60s}=16.1min[/tex]

Thus, the time required is, 16.1 minutes.

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