Easy heat transfer problem stuck

In summary, the open container holds 0.550 kg of ice at -15 Celsius and heat is supplied to it at a constant rate of 800 J/min for 500 min. After 21.7 minutes, the ice begins to melt. However, when the temperature rises above 0 Celsius is where the confusion lies. To calculate this, the total heat supplied (Qt) must be equal to the heat necessary for melting the ice (Q1) plus the heat of fusion of water (L) multiplied by the mass of the ice. After correcting a mistake in the calculation of the heat capacity of ice, the correct time for the temperature to rise above 0 Celsius is 44.7 minutes.
  • #1
Luis2101
13
0
An open container holds 0.550 kg of ice at -15 Celsius. The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 800 J/min for 500 min.

A) After how many minutes does the ice start to melt.
This part I calculated and got correct, it's 21.7 min.,

B) After how many minutes, from the time when the heating is first started, does the temperature begin to rise above 0 Celsius?
--
This is where I'm stuck. What I did was say that Qtotal would be equal to Q1 (the heat necessary in part A) + Heat of Fusion of water * the mass of the ice.
so: Qt = Q1 + Lm
L is 335,000 J/kg
m is 0.550 kg
Q1 from part A is 17325 J.

So, Qt = 35750 J

Time is Qt/rate of transfer = 35750 / 800 = 44.7 min, but that's incorrect...

What am I messing up?

-Luis
 
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  • #2
Luis2101 said:
So, Qt = 35750 J
You may wish to check this value.
 
  • #3
Yeah, I used the wrong value for the heat capacity of ice. (Used 33400 instead of 334000).
 

1. How can I determine the heat transfer coefficient for an easy heat transfer problem?

To determine the heat transfer coefficient, you will need to know the thermal conductivity of the materials involved, the surface area of heat transfer, and the temperature difference between the two materials. Once you have this information, you can use the equation h = kA/(LMTD), where h is the heat transfer coefficient, k is the thermal conductivity, A is the surface area, and LMTD is the logarithmic mean temperature difference.

2. What is the difference between conduction, convection, and radiation in heat transfer?

Conduction is the transfer of heat through a material by direct contact, convection is the transfer of heat through a fluid (such as air or water), and radiation is the transfer of heat through electromagnetic waves. In an easy heat transfer problem, you will likely encounter conduction as the primary mode of heat transfer.

3. How do I know if a heat transfer problem is considered "easy" or "difficult"?

An easy heat transfer problem typically involves a simple geometry, uniform material properties, and steady-state conditions. Difficult heat transfer problems may involve complex geometries, varying material properties, and transient conditions.

4. What are some common mistakes to avoid when solving an easy heat transfer problem?

Some common mistakes to avoid include forgetting to account for all sources of heat transfer, not considering the appropriate boundary conditions, and using incorrect equations or assumptions. It is important to carefully review your work and double-check your calculations to avoid these mistakes.

5. Are there any tips for solving easy heat transfer problems more efficiently?

One tip is to draw a heat transfer diagram to visualize the problem and identify all relevant variables and boundaries. Also, make sure to use appropriate units and conversions, and break down the problem into smaller, more manageable steps. It can also be helpful to work with a group or seek assistance from a professor or tutor if you are struggling with a particular problem.

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