Answer:
Heat absorbed = 1360.248 j
Explanation:
Given data:
Mass of brass = 298.3 g
Initial temperature = 30.0°C
Final temperature = 150°C
Specific heat capacity of brass = 0.038 J/g.°C
Heat absorbed = ?
Solution:
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
ΔT = 150°C - 30.0°C
ΔT = 120°C
Q = 298.3 g × 0.038 J/g.°C × 120°C
Q = 1360.248 j