Question

Suppose that 27 g of each of the following substances is initially at 28.0 ∘C. What is the final temperature of each substance upon absorbing 2.40 kJ of heat?
a) Gold
b) silver
c) aluminum
d) water

Answer 1

a)

Given:

Q = 2400 J

m = 27 g

C = 0.129 J/g.oC

Ti = 28 oC

use:

Q = m*C*(Tf-Ti)

2400.0 = 27.0*0.129*(Tf-28.0)

Tf -28.0 = 689 oC

Tf = 717 oC

Answer: 717 oC

b)

Given:

Q = 2400 J

m = 27 g

C = 0.24 J/g.oC

Ti = 28 oC

use:

Q = m*C*(Tf-Ti)

2400.0 = 27.0*0.24*(Tf-28.0)

Tf -28.0 = 370 oC

Tf = 398 oC

Answer: 398 oC

c)

Given:

Q = 2400 J

m = 27 g

C = 0.902 J/g.oC

Ti = 28 oC

use:

Q = m*C*(Tf-Ti)

2400.0 = 27.0*0.902*(Tf-28.0)

Tf -28.0 = 98.5 oC

Tf = 127 oC

Answer: 127 oC

d)

Given:

Q = 2400 J

m = 27 g

C = 4.184 J/g.oC

Ti = 28 oC

use:

Q = m*C*(Tf-Ti)

2400.0 = 27.0*4.184*(Tf-28.0)

Tf -28.0 = 21.2 oC

Tf = 49.2 oC

Answer: 49.2 oC