Question

An electric Range partner weighing 620 g is turned off after reaching temperature of 490.8 Celsius, and is allowed to cool down to 23.3 Celsius.

Calculate the specific heat of the burner if all the heat evolved from the burner is used to heat 579 g of water from 23.3 Celsius to 80.2 Celsius.

Answer 1

For water:

Given:

m = 579 g

C = 4.184 J/g.oC

Ti = 23.3 oC

Tf = 80.2 oC

use:

Q = m*C*(Tf-Ti)

Q = 579.0*4.184*(80.2-23.3)

Q = 1.378*10^5 J

This heat is lost by burner.

For burner:

Given:

Q = 137800 J

m = 620 g

Ti = 490.8 oC

Tf = 23.3 oC

use:

Q = m*C*(Tf-Ti)

137800.0 = 620.0*C*(23.3-490.8)

C = -0.4754 J/g.oC

Answer: -0.475 J/g.oC