Solving Heat Problems: 2 Questions & Exam Tomorrow

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In summary: A R2 bat has a thermal resistance of 2, a R3 has a thermal resistance of 3, and so on. A R4 bat has a thermal resistance of 4. To find out how much heat is lost through the roof in one day, you could use the thermal resistance of the material of the roof, multiply by the heat flow per square meter, and then scale it up to the room size and time period. ]There is a conductance [or its inverse resistance] which tells you the number of joules per second per square metre per degree difference each side. You need to cram that stuff now.[You may have seen ceiling insulation bats. They can
  • #1
shemer77
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Hi, I have 2 questions and an exam tomorrow, and I have no idea how my teacher answered these 2 problems, if someone can guide me through how to solve these problems i would really appreciate it. I am not looking for an answer just a step by step how to solve it and I am not really sure how to start.
How much heat is lost in one day through the roof of a 5.00m by 5.00m room?The room is maintained at a temperature of 20°C but the outdoor temperature is only -4°C.

How much heat, Q, is required to convert 0.500kg of ice at -30.0°C to steam at 140.0°C
For this one, i think my teacher would convert the ice to a liquid and then to a solid, and then taking taking some equation at each part, lol I am not really sure

thanks in advance :)
 
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  • #2
shemer77 said:
Hi, I have 2 questions and an exam tomorrow, and I have no idea how my teacher answered these 2 problems, if someone can guide me through how to solve these problems i would really appreciate it. I am not looking for an answer just a step by step how to solve it and I am not really sure how to start.

For this one, i think my teacher would convert the ice to a liquid and then to a solid, and then taking taking some equation at each part, lol I am not really sure thanks in advance :)

For the roof, I think you need to know the thermal resistance or thermal conductivity of the material of the roof. With that you can calculate the heat flow per square meter [Joules per sec per square meter] then scale it up for the room size and the time period.

For the ice to steam.

Heat the ice - what is the specific heat for ice? [it is 1 for water]
melt the ice - what is the latent heat for melting ice?
Heat the water
evaportate the water - what is the latent heat of evaportation?
heat the steam - what is the specific heat for steam?

NOTE: Suppose the latent heat for melting ice was 50 times the specific heat for water -it's NOT - the getting from ice at -10 to water at +10 would be the same as heating water from +5 to +75, ie 70 degrees.

You could convert each section to a "heating water" equivalent and do the final calculation in one go.

***Saying latent heat is 50 times specific heat I mean suppose you need X Joules to heat a mass of water by 1 degree but need 50X Joules to melt the same mass of ice.
 
  • #3
PeterO said:
For the roof, I think you need to know the thermal resistance or thermal conductivity of the material of the roof. With that you can calculate the heat flow per square meter [Joules per sec per square meter] then scale it up for the room size and the time period.

For the ice to steam.

Heat the ice - what is the specific heat for ice? [it is 1 for water]
melt the ice - what is the latent heat for melting ice?
Heat the water
evaportate the water - what is the latent heat of evaportation?
heat the steam - what is the specific heat for steam?

NOTE: Suppose the latent heat for melting ice was 50 times the specific heat for water -it's NOT - the getting from ice at -10 to water at +10 would be the same as heating water from +5 to +75, ie 70 degrees.

You could convert each section to a "heating water" equivalent and do the final calculation in one go.

***Saying latent heat is 50 times specific heat I mean suppose you need X Joules to heat a mass of water by 1 degree but need 50X Joules to melt the same mass of ice.


sorry, i didnt really learn this stuff. How would I go about calculating the heat flow?

and for the ice to steam, what do you mean by convert each section to a heating water equivalent?
 
  • #4
shemer77 said:
sorry, i didnt really learn this stuff. How would I go about calculating the heat flow?

and for the ice to steam, what do you mean by convert each section to a heating water equivalent?

If you don't immediately understand the conversions, ignore them and work it out the "standard" way.

There is a conductance [or its inverse resistance] which tells you the number of joules per second per square metre per degree difference each side. You need to cram that stuff now.

[You may have seen ceiling insulation bats. They can be R2, R3, R4 etc bats - that is referring to their thermal resistance.
 
  • #5


Hello,

Thank you for reaching out for help with your heat problems. I am happy to guide you through the process of solving these questions.

For the first question, we need to use the equation Q = mCΔT, where Q represents the amount of heat, m is the mass of the material, C is the specific heat capacity, and ΔT is the change in temperature. In this case, we are looking for the amount of heat lost, so we will use a negative value for Q.

First, we need to calculate the mass of the air in the room. Since the room is 5.00m by 5.00m, the area is 25m². Assuming the height of the room is 3m, the volume of air in the room is 25m² x 3m = 75m³. Air has a density of approximately 1.2 kg/m³, so the mass of air in the room is 75m³ x 1.2 kg/m³ = 90kg.

Next, we need to calculate the change in temperature. We know that the room is maintained at 20°C, and the outdoor temperature is -4°C. Therefore, the change in temperature is 20°C - (-4°C) = 24°C.

Now, we need to find the specific heat capacity of air. This value varies depending on the temperature and pressure, but for simplicity, we will use an average value of 1005 J/kg°C.

Putting all of this information into the equation, we get Q = 90kg x 1005 J/kg°C x (-24°C) = -2,160,000 J. This means that the room loses 2,160,000 joules of heat in one day through its roof.

For the second question, we need to use the equation Q = mL, where Q represents the amount of heat, m is the mass of the material, and L is the latent heat of fusion or vaporization. In this case, we are converting ice to steam, so we will use the latent heat of vaporization, which is 2,260,000 J/kg.

First, we need to calculate the mass of the ice. We are given that the mass is 0.500kg.

Next, we need to calculate the change in temperature. The ice is at -30.0°C, and we want to convert
 

FAQ: Solving Heat Problems: 2 Questions & Exam Tomorrow

What are the key steps to solving heat problems?

The key steps to solving heat problems include identifying the known and unknown variables, applying the appropriate formula or equation, and solving for the unknown variable. It is also important to check for unit consistency and make sure all units are the same.

How do I know which formula to use when solving a heat problem?

The formula used to solve a heat problem depends on the specific variables given and what is being asked to solve for. It is important to understand the principles of heat transfer and familiarize yourself with the different formulas for conduction, convection, and radiation.

What is the difference between conduction, convection, and radiation?

Conduction is the transfer of heat through a solid material, while convection is the transfer of heat through a fluid or gas. Radiation is the transfer of heat through electromagnetic waves. Each method of heat transfer has its own unique formula and variables.

Can I use the same formula for both heat and temperature problems?

No, heat and temperature are two different concepts and therefore require different formulas. Heat is a measure of energy, while temperature is a measure of the average kinetic energy of particles in a substance. However, the two are related and both play a role in heat transfer.

Are there any common mistakes to avoid when solving heat problems?

Some common mistakes to avoid when solving heat problems include using the wrong formula, not converting units correctly, and forgetting to include all relevant variables. It is also important to double check your calculations and make sure they are logical and reasonable.

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