Answer:
The energy absorbed when 6.20 grams of aluminum is heated from 10.0°C to 65.0°C is 306.9 Joules.
Explanation:
[tex]Q=mc\Delta T=mc\times (T_2-T_1)[/tex]
Where:
Q = heat absorbed or heat lost
c = specific heat of substance
m = Mass of the substance
ΔT = change in temperature of the substance
[tex]T_1,T_2[/tex] : Initial and final temperature of the substance
We have mass of aluminium = m = 6.20 g
Specific heat of aluminium= c = 0.900 J/g°C
Initial and final temperature of the aluminium [tex]T_1=10.0^oC [/tex]
Final temperature of the aluminium = [tex]T_2=65.0^oC[/tex]
Heat absorbed by the aluminium:
[tex]Q=6.20 g\times 0.900J/g^oC\times (65.0^oC-10.0^oC)=306.9 J[/tex]
The energy absorbed when 6.20 grams of aluminum is heated from 10.0°C to 65.0°C is 306.9 Joules.
