Answer:
Q = 1455 J
Explanation:
Given data:
Mass of block = 300 g
Initial temperature = 22.3 °C
Final temperature = 59.9°C
Specific heat of lead = 0.129 j/g.°C
Energy required = ?
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 = 59.9°C - 22.3 °C
ΔT = 37.6 °C
Q = 300 g × 0.129 j/g.°C × 37.6 °C
Q = 1455 J
