What is the sample's specific heat?

[Arejang]

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?

 

[hage567]

293.15 K is not the change in temperature.

 

[Moetasim]

Your calculation and approach seem correct. The specific heat of the sample is indeed 14.71 J/(kg*K). It is possible that the given answer is incorrect or that there was a mistake in the original problem. It is also possible that the sample is not a pure metal and has a different specific heat than expected. I would suggest checking with your instructor or a classmate to confirm the answer.