Help with a conservation of heat problem

In summary, the conversation discusses solving a specific heat problem involving a metal, water, and aluminum cup. The speaker shares their solution and asks for clarification on whether their method is correct and whether conductive heat transfer can only occur when a solid is present. They also confirm that a change in temperature in Kelvin is equal to a change in temperature in degrees Celsius.
  • #1
discordplus
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I already did it but its a different problem type that I really was unsure about so I just want to verify my solution.

Homework Statement


Problem: A 350-gram piece of metal at 100 C is dropped into a 100-gram aluminum cup containing 500 g of water at 15 C. The final temperature of the system is 40 C. What is the specific heat of the metal, assuming no heat is exchanged with the surroundings? Specific heat for aluminum is around 900 J/kg*K.

Basically:
m, m = .35kg
m, w = .5kg
m, a = .1kg
Ti, m = 100 C
Ti, a,w = 15 C
Tf = 40 C
Cp,w = 4186 J/kg*K
Cp,a = 900 J/kg*K
Cp,m = ?

Homework Equations


Q,a = -Q,b
delta Q = m*Cp*delta T

The Attempt at a Solution


how i did it:
Q,w,a = -Q,m
((4186 J/kg*K)(40-15)(.5) + (.1)(40-15)(900)) = -(.350)(C,m)(40-100)
52250 J + 2250 J = 21C,m
54500 = 21C,m
C,m = 2595.238 J/kg*K

Alright, step one here is really what I'm unsure about. I treated the aluminum and water as one part of the system, therefore I added the heat gained by the water and the aluminum as shown above, which I wasn't entirely sure of (in terms of that being allowed). Also I know a change in absolute temperature in K is the same as a change in temperature, deg C, but I was shaky on the interchangable use of kelvins and celsius degrees in the specific heats. I'm pretty sure that's valid, but I just would like to have that reaffirmed.

------

Anyways, that corresponded pretty closely with one of the choices, 2600 J/kg*deg K so I'm just making sure the work is right because usually I'm only working with two objects in the system, not three. Thanks in advance.

------

Also: Can conductive heat transfer only occur when a solid can mediate the energy transfer? I think this is true, but I'm not sure. I'm loose about the methods of heat transfer, very loose.
 
Last edited:
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  • #2
Welcome to PF.

Your method looks OK.

ΔT in °K = ΔT in °C

When the heat equalizes you know how much the one cooled down by the amount the others heated.
 
  • #3


Hello,

First of all, your method for solving the problem seems correct. You correctly identified the different masses and specific heats of the metal, water, and aluminum and used the appropriate equations to solve for the specific heat of the metal. As for your first question, it is perfectly valid to treat the water and aluminum as one part of the system and add their heat gains together. This is because they are both at the same temperature and will experience the same change in temperature, so their heat gains can be added together.

Regarding the use of kelvins and degrees Celsius, it is important to note that the specific heat values are defined in terms of kelvins (J/kg*K). However, when dealing with temperature changes, the change in temperature (delta T) can be expressed in either kelvins or degrees Celsius since they have the same magnitude. So in your calculation, it is valid to use 40-15=25 degrees Celsius or 40 K-15 K=25 K. It is also important to note that the specific heat values may vary slightly depending on the units used (J/g*K or J/kg*K), so it is always good to double check the units and make sure they are consistent.

To answer your last question, conductive heat transfer can occur between any two objects as long as they are in contact with each other. It does not necessarily require a solid to mediate the energy transfer, but solids are typically better conductors of heat compared to liquids or gases. However, in this specific problem, the heat transfer is occurring through conduction between the metal and the water/aluminum, so the metal acts as the mediator in this case.

Overall, your solution seems correct and your understanding of the concepts is good. Keep up the good work!
 

Related to Help with a conservation of heat problem

1. What is a conservation of heat problem?

A conservation of heat problem is a scenario in which the amount of heat entering and leaving a system is not balanced, resulting in a change in the overall temperature of the system. This can occur in various contexts, such as in homes and buildings, in industrial processes, or in natural environments.

2. How can I conserve heat in my home?

There are several ways to conserve heat in your home, such as insulating walls and attics, using weather stripping and caulk to seal air leaks, and using efficient heating systems and appliances. Additionally, simple habits like keeping doors and windows closed and using curtains can also help conserve heat.

3. What are some common causes of heat loss in a building?

Some common causes of heat loss in a building include poor insulation, air leaks through cracks and gaps, inefficient heating systems, and lack of ventilation. These factors can all contribute to an imbalance in the conservation of heat within a building.

4. How can I calculate the heat loss in a system?

To calculate the heat loss in a system, you will need to know the temperature difference between the inside and outside of the system, the surface area of the system, and the thermal resistance or conductivity of the materials involved. A heat loss calculator or the help of a professional can also aid in determining the heat loss in a system.

5. What are some strategies for conserving heat in natural environments?

In natural environments, heat conservation is important for maintaining the health and balance of ecosystems. Some strategies for conserving heat in these environments include preserving habitats and reducing human impact through sustainable practices, such as reducing carbon emissions and protecting natural resources.

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