The specific heat of aluminum is 0.90 J/ g*C degrees and of iron is 0.45 J/g*C. How much energy was absorbed by each 1-gram sample? The temperature of 1 gram of aluminum rises 5.6 C degrees and the temperature of 1 gram of iron rises 11.1 C degrees.
q = mass x specific heat x delta T.
For Al:
q = 1 g x 0.90 x 5.6 = ?
Follow the same process for Fe.
Thanks DrBob222!
To calculate the energy absorbed by each 1-gram sample, you can use the formula:
Energy = mass * specific heat * change in temperature
For the aluminum sample:
mass = 1 gram
specific heat of aluminum = 0.90 J/g*C
change in temperature = 5.6 C degrees
Energy = 1 * 0.90 * 5.6
Energy = 5.04 J
Therefore, the aluminum sample absorbed 5.04 Joules of energy.
For the iron sample:
mass = 1 gram
specific heat of iron = 0.45 J/g*C
change in temperature = 11.1 C degrees
Energy = 1 * 0.45 * 11.1
Energy = 5.60 J
Therefore, the iron sample absorbed 5.60 Joules of energy.
To calculate the energy absorbed by each 1-gram sample of aluminum and iron, we can use the formula:
Energy = mass * specific heat * change in temperature
For the aluminum sample:
Energy = 1g * 0.90 J/g°C * 5.6°C = 5.04 J
For the iron sample:
Energy = 1g * 0.45 J/g°C * 11.1°C = 4.995 J
Therefore, the energy absorbed by the 1-gram sample of aluminum is 5.04 J, and the energy absorbed by the 1-gram sample of iron is 4.995 J.