5. How much heat (in kJ) is given out when 85.0 g of lead cools from 200.0
°C to 10.0°C? (Cp of lead = 0.129 J/g °C)

2.08335 kj

Explanation:

Given data:

Mass of lead = 85 g

Initial temperature = 200°C

Final temperature = 10°C

Heat lost  = ?

Cp of lead = 0.129 j/g.°C

Solution:

Specific heat capacity:  Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 10°C - 200°C

ΔT = -190°C

Now we will put the values:

Q = 85 g × 0.129 j/g.°C ×  -190°C

Q = 2083.35 j

Joule to kilo joule:

2083.35/1000 = 2.08335 kj