Question

A sample of cobalt, A, with a mass of 5.00 g, is initially at 25.0 °C. When this sample gains 6.70 J of heat, the temperature rises to 27.9 °C. Another sample of cobalt, B, with a mass of 7.00 g, is initially at 25.0 °C. If sample B gains 5.00 J of heat, what is the final temperature of sample B. (Hint: think about the specific heat of both samples.)

Answer 1

\( q = mC\Delta T\)

q is heat
m is mass
C is the specific heat
\(\Delta T\) is the change in temperature

Answer 2

So to solve your question, let's start with sample A

Sample A of Cobalt

Mass = 5 g
Initial temperature = 25.0 °C
Final temperature = 27.9 °C
Heat = 6.70 J
C = specific heat = ???

Use the formula I provided to you

Remember \(\Large \Delta T = T_{final} - T_{initial}\)

Once we calculate the specific heat of cobalt, we can use this specific heat (since it's a constant for the same material) and use it when we're solving for the final temperature of sample B

Answer 3

Since you're offline, I'll give you the steps for the rest of the problem as well

So once you find the specific heat of Cobalt, we're going to use it now for sample B

So Sample B of Cobalt

Mass = 7 g
Initial temperature = 25.0 °C
Final temperature = ???
Heat = 5 J
and specific heat = whatever you calculated earlier

Now put all the numbers into the equation

\( q = mC\Delta T \)

another way to write it would be

\(\Large q = mC(T_{final} - T_{initial}) \)

you'll be solving for \(\large T_{final}\) and you have all the other values so it should be pretty straightforward :)