Answer:
T initial = 19.55°C
Explanation:
Given data:
Mass of Al = 35 g
Energy produces = 781.2 J
Final temperature = 44.3°C
Initial temperature = ?
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 Al is 0.902 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 = Final temperature - initial temperature
Q = m.c. ΔT
781.2 J = 35 g×0.902 J/g.°C× (T final - T initial)
781.2 J = 35 g×0.902 J/g.°C× (44.3°C -T initial )
781.2 J = 31.57 j/°C× (44.3°C -T initial )
781.2 J/ 31.57 j/°C = (44.3°C -T initial )
24.75°C = (44.3°C -T initial )
24.75° - 44.3°C = -T initial
T initial = 19.55°C
