Question

1) Calculate the temperature change for mercury if 160 g of the metal absorbs 1500 J of heat energy. Mercury's specific heat is 0.14 J/g°C.

2) How much heat is given out when 85 g of lead cools from 200.0°C? The specific heat of lead is 0.11 J/g°C

3) What is the final temperature of a substance if 1404 J of energy is added to 38 g of it at 26°C? The specific heat is 0.24 J/g°C

Answer 1

1. 66.96°C

2. -1776.5J

3. 179.9°C

Step-by-step explanation:

Using the formula;

Q = m × c × ∆T

Where;

Q = amount of heat (J)

m = mass (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

QUESTION 1:

∆T = ?

m = 160g

c = 0.14 J/g°C

Q = 1500 J

1500 = 160 × 0.14 × ∆T

1500 = 22.4∆T

∆T = 1500/22.4

∆T = 66.96°C

QUESTION 2:

Q = ?

m = 85g

c = 0.11 J/g°C

∆T = 10°C - 200°C = -190°C

Q = 85 × 0.11 × -190

Q = -1776.5J

QUESTION 3:

Q = 1404 J

m = 38g

c = 0.24 J/g°C

Final temperature = ?

Initial temperature = 26°C

1404 = 38 × 0.24 × (T2 - 26)

1404 = 9.12 (T2 - 26)

1404 = 9.12T2 - 237.12

9.12T2 = 1404 + 237.12

9.12T2 = 1641.12

T2 = 1641.12 ÷ 9.12

T2 = 179.9°C