Answer:
Q = -1045 J
Explanation:
Given data:
Mass of water = 5.00 g
Initial temperature = 348.0 K
Final temperature = 298.0 K
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.K
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 = 298.0 K - 348.0 K
ΔT = - 50 K
Q = 5.0 g ×4.18 J/g.K× - 50 K
Q = -1045 J
