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Problem 2. Electric Heating of a Room An air-tight room contains 80 kg of air, and a 2-kW baseboard electric resistance heater in the room is turned on and kept on for 15 min. Calculate the temperature rise of air at the end of 15 min.

Respuesta :

Remzwisdom

Answer:

The rise in temperature after 15mins is (ΔT)=[tex]31.34[/tex]°c

Explanation:

The solution is obtained from energy balance;

W=M x Cv x ΔT

where;

W(workdone)=quantity of energy in joules

Workdone=Power x Time

Power=2kw

Time=15mins=15 x 60=900secs

Workdone=2 x 900=1800KJ

Cv=Specific heat capacity of air= 0.718 kJ/kg.K

M=Mass of air=80kg

ΔT= change in temperature in °C

therefore;

ΔT=[tex]\frac{W}{Cv *M}[/tex]

ΔT=[tex]\frac{1800}{0.718*80}[/tex]

ΔT=[tex]\frac{1800}{57.44}[/tex]

ΔT=[tex]31.34[/tex]°c

fahadisahadam

Answer:

The temperature rise of air at the end of 15 min is [tex]31.34[/tex]°C

Explanation:

Data

[tex]t=15min, t=15*60sec, t=900seconds[/tex]

[tex]W=2kW[/tex]

[tex]m=80kg[/tex]

[tex]constant c_{v}[/tex]=0.718

We can use energy balance to solve the problem.

The energy balance stated that:

[tex]W=[/tex]Δ[tex]U[/tex]

[tex]W=mc_{v}[/tex]Δ[tex]T[/tex]

[tex]W=\frac{m}{t} c_{v}[/tex]Δ[tex]T[/tex]

we can now make ΔT the subject of the formula

ΔT[tex]=\frac{W*t}{mc_{v} }[/tex]

substitute the values

ΔT[tex]=\frac{2*900}{80*0.718}[/tex]

ΔT=[tex]=31.34[/tex]°C