Answer:
Q = 5227.5 J
Explanation:
Given data:
Mass of water = 50.0 g
Temperature increase by = ΔT = 25°C
Specific heat capacity of water = 4182 J/Kg.°C
Heat required = ?
Solution:
(50.0 g×1 Kg/1000 g=0.05 Kg)
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
Q = 0.05 Kg ×4182 J/Kg.°C × 25°C
Q = 5227.5 J
