4. A total of 226 J of heat are absorbed as 58.3 g of lead is heated from 12.0°C to 42.0°C. From this data, what is the specific heat of lead?||
I

Specific heat capacity of lead is  0.129 J/g.°C

Explanation:

Given data:

Mass of lead = 58.3 g

Heat absorbed = 226 J

Initial temperature = 12.0°C

Final temperature = 42.0°C

Specific heat capacity of lead = ?

Solution:

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 = 42.0°C - 12.0°C

ΔT = 30°C

226 J = 58.3 g × c × 30°C

226 J = 1749 g.°C × c

c = 226 J /1749 g.°C

c = 0.129 J/g.°C

Specific heat capacity:

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

Specific heat capacity of lead is  0.129 J/g.°C

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