Thermodynamics - Hypothermia - Heat loss

[Jess_18033152]

Homework Statement

A careless farmer sends this sheep out to graze on a very cold winter day when the temperature is -10oC. The sheep's coat has a thickness of 4.7 cm and a surface area 1.3 m2.

Calculate the rate of heat loss from the sheep on this cold day. Assume that heat is lost only through the sheep's wool coat.

Homework Equations

Sheep core body temp = 39 C
Sheep develop hypothermia if body temp falls below 37 C
thermal conductivity (k) of sheep's wool = 0.038 W/m/K
specific heat capacity (c) of sheep = 4180 J/kg/K

mass of sheep = 63 kg
Q = 527000 J

delta Q/ delta t = - kA (delta T/ delta x)

What number should we use to calculate delta T?
We were given 3 temperatures;
- Sheep core body temp = 39 C
- Sheep develop hypothermia if body temp falls below 37 C
- atmospheric temperature = -10 C

 

[Orodruin]

What number should we use to calculate delta T?

What do you think? What is the definition of $\Delta T$? Does it matter much which of the sheep temperatures you use?

Is there a second part of the question that asks how long it takes for the sheep to develop hypothermia? Please state the entire problem even if you are only asking for the first part.

 

[Jess_18033152]

What do you think? What is the definition of $\Delta T$? Does it matter much which of the sheep temperatures you use?

Is there a second part of the question that asks how long it takes for the sheep to develop hypothermia? Please state the entire problem even if you are only asking for the first part.

Hi :), thank you for your help. No there is not a second question relating to how long it takes for the sheep to develop hypothermia.

Also, with the change in temp (delta T) which of the following calculations would be correct?
delta T = -10 + 39 = 29 C
or
delta T = 39 + 10 = 49 C

?

 

[Orodruin]

$\Delta T$ is the temperature difference. Which of your expressions is the difference between the sheep and outside temperature?

 

[Jess_18033152]

$\Delta T$ is the temperature difference. Which of your expressions is the difference between the sheep and outside temperature?

Thank you :), got the question right... it was 39 + 10 = 49 C

 

[Jess_18033152]

Thank you :), got the question right... it was 39 + 10 = 49 C

Sorry again, found that there was a question about calculating how long it will take this sheep to develop hypothermia if left outside on this cold day. With the final answer in hours. Same numbers given as shown above at the start of my original forum.

I have attempted the question, and told my answer was wrong using the following equation;

t = Q x L/ k x A x delta T
where
Q = 51.5 W
L is assumed to be x = 0.047 m
k = 0.038 W/m/K
A = 1.3 m^2
delta T = 49 C

t = (51.5 x 0.047)/(0.038 x 1.3 x 49)
t = 0.9999586879
= 1.0 s (3sf)

which converts to 0.000278 hours and was found to be incorrect.

I was then told;
You'll need to use your answers to the previous two questions (4 and 5) to answer this question.
4) Q = m c delta T
= 63 kg (mass of sheep) x 4180 x (39 - 37)
= 527000 J
5) being the question I originally asked in this forum;
Q/ delta t = 51.5 W

You can assume that the rate of heat transfer does not significantly change as the sheep cools down by 2oC.

Where do i need to go with this question so I can use both answer from question 4 and question 5?

 

[Orodruin]

Q = m c delta T

Note that this is a different $\Delta T$ from your previous task. In this case it is the change in temperature of the sheep, not the difference in temperatures. $Q/\Delta t = 51.5$ W is just an expression for the average power needed to remove heat Q in time $\Delta t$. Note that $Q$ here is the total heat, not the heat loss rate as you had in your first part.

Also note that the problem seems to be treating the sheep as a non-living thing. In reality, the sheep will also heat up due to its internal metabolism. For example, a human body at rests produces around 100 W of heat.

 

[Jess_18033152]

Note that this is a different $\Delta T$ from your previous task. In this case it is the change in temperature of the sheep, not the difference in temperatures. $Q/\Delta t = 51.5$ W is just an expression for the average power needed to remove heat Q in time $\Delta t$. Note that $Q$ here is the total heat, not the heat loss rate as you had in your first part.

Also note that the problem seems to be treating the sheep as a non-living thing. In reality, the sheep will also heat up due to its internal metabolism. For example, a human body at rests produces around 100 W of heat.

Is this how I would calculate it then?

t = Q x L/ k x A x delta T
= (51.5 x 0.047)/(0.038 x 1.3 x 2)
= 24.5 s

converted to hours
= 0.00681 hours?

 

[Orodruin]

Is this how I would calculate it then?

t = Q x L/ k x A x delta T
= (51.5 x 0.047)/(0.038 x 1.3 x 2)
= 24.5 s

No. What you are computing here does not even have the correct physical dimension! If you follow the units - which you should always do! - they will not come out to be a number of seconds, it will come out dimensionless! All you have done here is to take the ratio between the temperature difference between the sheep and the outside and the temperature change in the sheep. You need to follow the steps you were told.

converted to hours
= 0.00681 hours?

Does this seem reasonable to you? Do you think you would develop hypothermia in less than a minute? And you do not have a coat of wool.

 

[Jess_18033152]

No. What you are computing here does not even have the correct physical dimension! If you follow the units - which you should always do! - they will not come out to be a number of seconds, it will come out dimensionless! All you have done here is to take the ratio between the temperature difference between the sheep and the outside and the temperature change in the sheep. You need to follow the steps you were told.Does this seem reasonable to you? Do you think you would develop hypothermia in less than a minute? And you do not have a coat of wool.

Sorry, but I have absolutely no idea where to go from this...

 

[Orodruin]

What about the proposed approach is unclear?

 

[Jess_18033152]

What about the proposed approach is unclear?

Im not sure whether the equation I'm using is right or whether I need to use a new equation and what given numbers I should be using?

 

[Orodruin]

I think you are focusing too much on inserting things into equations rather than understanding what the equations are telling you. This approach is going to come back to bite you in the end. This is particularly true when, as here, you have taken Q to mean different things in different parts of the problem and you are not careful with the units you are inserting.

Can you try to describe, in words, what these equations are telling you?
$$
\Delta Q = mc \Delta T, \quad P = \frac{\Delta Q}{\Delta t}
$$