Heat loss problem- thermodynamics

[mmoadi]

Homework Statement

We are heating up the cabin log with a furnace that has strength of 4 kW. The surface of the walls of the log cabin is 35 m². What is the temperature in the log cabin, if the outside temperature is -20 °C? The thermal conductivity of wood is 0.4 W/mK; the average wall thickness is 15 cm. How many degrees will be the temperature reduced in the log cabin, if we install windows into walls, through which escapes heat flow of 1800 W, and the surface area of the windows is 10 m²?

Homework Equations

ΔQ/Δt= K [AΔT/d]

The Attempt at a Solution

First part: What is the temperature in the log cabin, if the outside temperature is -20 °C?

ΔQ/Δt= K [A(T(hot)-T(cold))/d]
4000 W= 0.4 W/mK [T(hot)-253 K)/ 0.15 m
T(hot)= 295.86 K= 22.7 C

Are my calculations correct?

Now, for the second part, I really don’t know how to approach it. I tried several times but my results didn’t make any sense (either too high or too low temperature). Can someone please help me?
Thank you!

 

[Andrew Mason]

Homework Statement

We are heating up the cabin log with a furnace that has strength of 4 kW. The surface of the walls of the log cabin is 35 m². What is the temperature in the log cabin, if the outside temperature is -20 °C? The thermal conductivity of wood is 0.4 W/mK; the average wall thickness is 15 cm. How many degrees will be the temperature reduced in the log cabin, if we install windows into walls, through which escapes heat flow of 1800 W, and the surface area of the windows is 10 m²?

Homework Equations

ΔQ/Δt= K [AΔT/d]

The Attempt at a Solution

First part: What is the temperature in the log cabin, if the outside temperature is -20 °C?

ΔQ/Δt= K [A(T(hot)-T(cold))/d]
4000 W= 0.4 W/mK [T(hot)-253 K)/ 0.15 m
T(hot)= 295.86 K= 22.7 C

Are my calculations correct?

Looks good to me.

Now, for the second part, I really don’t know how to approach it. I tried several times but my results didn’t make any sense (either too high or too low temperature). Can someone please help me?

It is the same calculation except that you have effectively 1800 W less heat (ie. 2200 W) from the furnace and 10m^2 less wall area for the 2200 W to conduct through.

AM

 

[mmoadi]

<sub>Second part: How many degrees will be the temperature reduced in the log cabin, if we install windows into walls, through which escapes heat flow of 1800 W, and the surface area of the windows is 10m²?

ΔQ/Δt= K [A(T(hot)-T(cold))/d]
2200 W= 0.4 W/mK [25 m² (T(hot)-253 K))/ 0.15 m]
T(hot)= 286 K= 12.9 ºC

The temperature in the room is reduced for: ΔT= 22.7 ºC - 12.9 ºC= 9.8 ºC

Are my calculations correct?
Thank you for helping!</sub>