Answer:
Q = 336000 j
Explanation:
Specific heat capacity:
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
Given data:
Specific heat capacity = 4200 j/Kg °C
Mass of water = 1 Kg
Initial temperature = 20°C
Final temperature = 100°C
Heat required = ?
Solution:
ΔT = T2 - T1
ΔT = 100°C - 20°C
ΔT = 80 °C
Q = m.c. ΔT
Q = 1 kg ×4200 j/Kg °C ×80 °C
Q = 336000 j
