Calculating Specific Heat of a Tin Sample: A Quick Question and Solution

  • Thread starter Aikenfan
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In summary, the problem involves a 250 gram piece of tin at 99 degrees C being dropped into 100 grams of water at 10 degrees C. The final temperature is 20 degrees C. To calculate the specific heat of the tin sample, we use the formula q = mcΔT and the specific heat of water, c = 4.186 KJ/kg.K. After converting to Kelvin, the temperatures are 372.15K, 293.15K, and 283.15K. To solve for x, we use the values and the equation 0.250 * x * (372.15 - 293.15) = 0.100 * 4.186 * (293.
  • #1
Aikenfan
48
0
I saw this problem online...

A 250 gram piece of tin at 99 degrees C. is dropped into 100 grams of water at 10 degress C. If the final temperature is 20 degrees C., calculate the specific heat of the tin sample. Be sure to include the appropriate units !
Note c for water = 4.186 KJ/kg.K

Solution 41766

Assume K = 273.15 + C where K is Kelvin and C is Celsius.
Hints:
Let the specific heat of tin be x kJ/(kg*K).
0.250 * x * (372.15 - 293.15) = 0.100 * 4.186 * (293.15 - 283.15)
Solve for x.

and i just wondered...where did the (372.15 - 293.15) come from? thank you very much!
 
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  • #2
what are the 3 temps after converting to Kelvin?

then heat lost by tin=heat rise of water
 
  • #3
99--- - 174.15
10---- -263.15
20---- -253.15
 
  • #4
you sure, k=273.15+C
 

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