Answer:
Q = -14322.77 J
Step-by-step explanation:
Given data:
Mass of water = 55.0 g
Initial temperature = 87.3°C
Final temperature = 25.0 °C
Heat given off = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water is 4.18 J/g.°C
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 = 25.0 °C - 87.3°C
ΔT = - 62.3 °C
Q = 55.0 g×4.18 J/g.°C × - 62.3 °C
Q = -14322.77 J