Heat loss problem- thermodynamics

In summary, the log cabin is heated by a 4 kW furnace, with a surface area of 35 m² and an outside temperature of -20 °C. The temperature inside the cabin is 22.7 °C. If windows with a heat flow of 1800 W and a surface area of 10 m² are installed, the temperature inside the cabin will be reduced by 9.8 °C to 12.9 °C. The thermal conductivity of wood is 0.4 W/mK and the average wall thickness is 15 cm. The calculations for both parts appear to be correct.
  • #1
mmoadi
157
0

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!
 
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  • #2
mmoadi said:

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
 
  • #3
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!
 

Related to Heat loss problem- thermodynamics

1. What is heat loss and why is it a problem in thermodynamics?

Heat loss is the transfer of thermal energy from a system to its surroundings. In thermodynamics, it is a problem because it can decrease the efficiency of a system and lead to a decrease in the system's overall performance.

2. What are the main factors that contribute to heat loss?

The main factors that contribute to heat loss are the temperature difference between the system and its surroundings, the surface area of the system, and the thermal conductivity of the materials involved.

3. How can heat loss be minimized in a thermodynamic system?

Heat loss can be minimized by using materials with low thermal conductivity, increasing the insulation of the system, and reducing the surface area exposed to the surroundings. Additionally, maintaining a smaller temperature difference between the system and its surroundings can also help minimize heat loss.

4. What are the consequences of excessive heat loss in a thermodynamic system?

Excessive heat loss can lead to a decrease in the system's efficiency, which can result in higher energy costs and a decrease in the system's overall performance. It can also cause temperature fluctuations and potential damage to the system.

5. How is heat loss calculated and measured in thermodynamics?

Heat loss is typically calculated using the equation Q = U x A x (T1-T2), where Q is the heat loss, U is the thermal conductivity, A is the surface area, and T1 and T2 are the temperatures of the system and its surroundings, respectively. It can be measured using various methods such as thermocouples, heat flux sensors, and infrared cameras.

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