What is the sample's specific heat?

  • Thread starter Arejang
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Homework Statement


You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is 30.0 N. You carefully add 1.30×10^4 J of heat energy to the sample and find that its temperature rises 20.0 degrees C.

What is the sample's specific heat?


Homework Equations



Q=mc (delta)T

The Attempt at a Solution



I solved for c giving:

c=Q/(m*(delta)T)

I converted the given degrees Celsius to Kelvin by adding 273.15, and converted the 30 Newtons to kg by dividing by 9.8 m/s^2. The resulting numbers, 293.15 was used for delta T, and 2.90 kg was used for m.

This was my resultant value and calculation prior to plug and chug:

1.30*10^4 J/(2.90 kg*293.15 K)

Plug and chug gave me 14.71 J/(kg*K). But it's not the correct answer. Could anyone show me what I did wrong?
 

Answers and Replies

  • #2
hage567
Homework Helper
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293.15 K is not the change in temperature.
 

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