Question

A)In the laboratory a student uses a "coffee cup" calorimeter to determine the specific heat of a metal.

She heats 19.3 grams of chromium to 98.47°C and then drops it into a cup containing 81.8 grams of water at 23.17°C. She measures the final temperature to be 24.97°C.

Assuming that all of the heat is transferred to the water, she calculates the specific heat of chromium to be __________________ J/g°C.

B) An electric range burner weighing 616.0 grams is turned off after reaching a temperature of 490.2°C, and is allowed to cool down to 23.1°C.

Calculate the specific heat of the burner if all the heat evolved from the burner is used to heat 552.0 grams of water from 23.1°C to80.9°C.


Answer _________________J/g°C

Answer 1

For PArt A and B use the following expression: Q1 = Q2 ---> Q = m*cp*(Tf-Ti)

Now in part A, with the specific heat of water (4.184 J/g°C) and the data, let's calculate the heat:
Q = 81.8 * 4.184 * (23.17-24.97) = -616.05 J

Now, to calculate cp = Q/m(Tf-Ti)
cp = -616.05 / 19.3 (24.97-98.4)
cp = 0.4347 J/g°C

For part B, you'll have to do exactly the same but changing values. With the data of water calculate the heat, and then solve for cp and you should get the final result. I will leave this part to you for better understanding.

Hope this helps