Answer:
Q = 57934.8 J
Explanation:
Given data:
Mass of water = 180.0 g
Initial temperature = 19°C
Fina temperature = 96°C
Specific heat capacity of water = 4.18 J/g.°C
Heat absorbed = ?
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 = 96°C - 19°C
ΔT = 77°C
Q = 180.0 g×4.18 J/g.°C×77°C
Q = 57934.8 J
