Answer:
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
