Answer:
Q = 813.12 j
Explanation:
Given data:
Mass of lead = 21 g
Initial temperature = 25°C
Final temperature = 327.5°C
Amount of heat absorbed by 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 = T2 - T1
ΔT = 327.5°C - 25°C
ΔT = 302.5°C
Specific heat capacity of lead = 0.128 j/g.°C
Q = m.c. ΔT
Q = 21 g. 0.128 j/g.°C.302.5°C
Q = 813.12 j
