a 1500 - w heater is designed to be plugged into a 120 - v outlet. part c how long does it take to raise the…

a 1500 - w heater is designed to be plugged into a 120 - v outlet. part c 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 1.20 kg/m³. express your answer numerically in minutes using three significant figures. view available hint(s)

a 1500 - w heater is designed to be plugged into a 120 - v outlet. part c 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 1.20 kg/m³. express your answer numerically in minutes using three significant figures. view available hint(s)

Answer

Explanation:

Step1: Calculate the volume of the room

$V = 3.00\times5.00\times8.00=120\ m^{3}$

Step2: Calculate the mass of the air

Using $\rho=\frac{m}{V}$, we get $m=\rho V$. Substituting $\rho = 1.20\ kg/m^{3}$ and $V = 120\ m^{3}$, we have $m=1.20\times120 = 144\ kg$

Step3: Calculate the heat required

Using $Q = mc\Delta T$, with $c = 1006\ J/(kg\cdot^{\circ}C)$, $m = 144\ kg$ and $\Delta T=10.0^{\circ}C$, we get $Q=144\times1006\times10.0 = 1.44864\times10^{6}\ J$

Step4: Calculate the time

The power of the heater is $P = 1500\ W$. Since $P=\frac{Q}{t}$, then $t=\frac{Q}{P}$. Substituting $Q = 1.44864\times10^{6}\ J$ and $P = 1500\ W$, we have $t=\frac{1.44864\times10^{6}}{1500}=965.76\ s$. Converting to minutes, $t=\frac{965.76}{60}\approx16.1\ min$

Answer:

$16.1$