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Finding Specific Heat of Metal

  • Thread starter eoneil
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  • #1
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Homework Statement


A piece of metal of weight 50 grams is heated to a temperature of 100°C and then introduced into a calorimeter containing water at 14°C, the maximum temperature reached being 20°C. If the weight of the calorimeter is 5 grams, the specific heat of the metal of which it is made is 0.095 and the weight of the water it contains is 92 grams, calculate the specific heat of the given metal.



Homework Equations


deltaH=cmdeltaT

The Attempt at a Solution



First the metal is added to a known mass of water at a known temperature (50g added to water at 100 degrees) in a calorimeter. The metal and the water eventually reach thermal equilibrium (Te), whereby the metal cools from 100, the water warms from 14. Since heat energy is preserved, heat lost by the metal equals the heat gained by the water (assuming the calorimeter absorbs none of the heat).

The specific heat can be given as the calories per gram of material per degree change. The heat loss of the metal is:

Heat Lost = (specific heat)(50g)(100 C – Te)
Specific heat of water= 1.00cal/g deg
Heat Gained = (1.00cal/g deg)(92g)( Te – 14 C)
Heat Lost = Heat Gained, and the final temperature, Te, can be measured
With Te= 20 C.
Specific heat)(50g)(86 C)= (1.00cal/g deg)(92g)(6 C)
Specific heat=0.1283 cal/g deg

Is this right?
 

Answers and Replies

  • #2
gneill
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I don't see where you've accounted for the heat associated with the calorimeter. What are the units for the given value of specific heat of the calorimeter metal?
 
  • #3
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I'm confused as to where to include specific heat of the calorimeter metal, 0.095. And is it included alongside the net weight of 92g water + 5g calorimeter?
 
  • #4
gneill
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Presumably the calorimeter is the container holding the water. So it should participate.

Did the original problem not specify the units for the specific heat, or just the bare number, 0.095? Are we supposed to assume cal/gm-K ? J/kg-K? Something else?
 
  • #5
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I agree that the calorimeter values should be incorporated. As for the specific heat given, no units included.
 
  • #6
gneill
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I agree that the calorimeter values should be incorporated. As for the specific heat given, no units included.
So, I suppose the best thing to do is assume the same units as those already used in the problem, cal/gm-K.
 
  • #7
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I think the answer lies somewhere in the 'heat gained' step.

(Specific heat)(50g)(80 C)= (1.00cal/g deg)(92g)(6 deg)

Since the water sits in the calorimeter, it must be at the same temperature. The calorimeter heats up alongside the water, but their specific heats are different. How do I combine two specific heats?
 
  • #8
gneill
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20,793
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Since the water sits in the calorimeter, it must be at the same temperature. The calorimeter heats up alongside the water, but their specific heats are different. How do I combine two specific heats?
I usually find it easier to combine the heat capacities, i.e., C = m1*cp1 + m2*cp2. you end up with a combined object with mass M = m1 + m2 and effective heat capacity C/M.

In fact, you can do the whole problem by calculating the total heat of all the components before mixing (use absolute temperature scale), and the total heat capacity when everything is together. The final temperature should equal the total heat over the total heat capacity: Tf = Q/C. Your unknown value (cp of the unknown metal) will appear in the final expression. Solve for it.
 

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