Question

A 38.8 gram piece of metal absorbs 181J as it temperature increases from 25.0 degree celsius to 36.0 degree celsius. What is the specific heat of the metal?
__________________________J/g degree C

Answer 1

Answer: 0.424 J/g°C

Explanation:

For this problem, we would have to manipulate the equaiton for heat, q=mCT. Specific heat is the C in the equation. Since we are looking for specific heat, we manipulate the equation so that it says C=.

$C=\frac{q}{m(deltaT)}$

*I didn't know how to type in delta so I just wrote the word delta, but pretend you see a Δ.

Now that we have our equation, we can plug in our values and solve.

$C=\frac{181J}{(38.8g)(36-25°C)}$

*Please ignore the capital A in the equation. It pops up every time I type in the ° sign.

$C=0.424J/g°C$