Question

Suppose that 27 g of each substance is initially at 29.0 ∘C.

What is the final temperature of gold upon absorbing 2.25 kJ of heat?

What is the final temperature of silver upon absorbing 2.25 kJ of heat?

What is the final temperature of aluminum upon absorbing 2.25 kJ of heat?

What is the final temperature of water upon absorbing 2.25 kJ of heat?

Answer 1

1)

Given:

Q = 2250 J

m = 27 g

C = 0.129 J/g.oC

Ti = 29 oC

use:

Q = m*C*(Tf-Ti)

2250.0 = 27.0*0.129*(Tf-29.0)

Tf -29.0 = 646 oC

Tf = 675 oC

Answer: 675 oC

2)

Given:

Q = 2250 J

m = 27 g

C = 0.24 J/g.oC

Ti = 29 oC

use:

Q = m*C*(Tf-Ti)

2250.0 = 27.0*0.24*(Tf-29.0)

Tf -29.0 = 347 oC

Tf = 376 oC

Answer: 376 oC

3)

Given:

Q = 2250 J

m = 27 g

C = 0.902 J/g.oC

Ti = 29 oC

use:

Q = m*C*(Tf-Ti)

2250.0 = 27.0*0.902*(Tf-29.0)

Tf -29.0 = 92.4 oC

Tf = 121 oC

Answer: 121 oC

4)

Given:

Q = 2250 J

m = 27 g

C = 4.184 J/g.oC

Ti = 29 oC

use:

Q = m*C*(Tf-Ti)

2250.0 = 27.0*4.184*(Tf-29.0)

Tf -29.0 = 19.9 oC

Tf = 48.9 oC

Answer: 48.9 oC