Question

An electric range burner weighing 616.0 grams is turned off after reaching a temperature of 465.4°C, and is allowed to cool down to 23.2°C. Calculate the specific heat of the burner if all the heat evolved from the burner is used to heat 575.0 grams of water from 23.2°C to 81.4°C. Answer O /gºC Submit Answer Retry Entire Group 8 more group attempts remaining

Answer 1

Step 1: calculate heat absorbed by water

Given:

m = 575 g

C = 4.184 J/g.oC

Ti = 23.2 oC

Tf = 81.4 oC

use:

Q = m*C*(Tf-Ti)

Q = 575.0*4.184*(81.4-23.2)

Q = 140017 J

Step 2: calculate specific heat of burner

The above heat is released by burner

Given:

Q = -140017 J

m = 616 g

Ti = 465.4 oC

Tf = 23.2 oC

use:

Q = m*C*(Tf-Ti)

-140017 = 616.0*C*(23.2-465.4)

C = 0.514 J/g.oC

Answer: 0.514 J/g.oC