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

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To calculate the specific heat of a tin sample, a 250-gram piece of tin at 99 degrees Celsius is dropped into 100 grams of water at 10 degrees Celsius, resulting in a final temperature of 20 degrees Celsius. The specific heat of tin is represented as x kJ/(kg*K), and the equation used is 0.250 * x * (372.15 - 293.15) = 0.100 * 4.186 * (293.15 - 283.15). The temperatures in Kelvin are derived by adding 273.15 to the Celsius values, leading to 372.15 K for tin at 99 degrees C, 283.15 K for water at 10 degrees C, and 293.15 K for the final temperature. The discussion clarifies the conversion to Kelvin and the heat transfer principles involved in the calculation. Understanding these conversions is essential for accurate specific heat calculations.
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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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what are the 3 temps after converting to Kelvin?

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

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