Solve the following problem: Heat is converted into Work and vice versa, that causes change in internal energy. 1 cal. = 4.184 Joule (J) - is the S.I. unit for energy Formula: AU = Q-W Q=W+AU
1. If 200 J of energy is added to a system when no external work has been done, by how much will the thermal energy of the system be raised?
2. A 100 J of energy is added to a system that does 10 J of external work. By how much is the thermal energy of the system raised?
3. A beaker containing water has 1500 J of work
done on it by stirring and 200 cal of heat
added to it from a hot plate.
What is the change in the internal energy of the
water in Joules?
b. Determine the change in the internal energy of the water in calories.
1. If 200 J of energy is added to a system when no external work has been done, by how much will the thermal energy of the system be raised?
2. A 100 J of energy is added to a system that does 10 J of external work. By how much is the thermal energy of the system raised?
3. A beaker containing water has 1500 J of work
done on it by stirring and 200 cal of heat
added to it from a hot plate.
What is the change in the internal energy of the
water in Joules?
b. Determine the change in the internal energy of the water in calories.
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Answer:
1. The increase in thermal energy will be 200 J.
2. The change in internal energy (AU) can be calculated using the formula AU = Q - W. Plugging in the values we get, AU = 100 J - 10 J = 90 J. Therefore, the thermal energy of the system will be raised by 90 J.
3. The change in internal energy (AU) can be calculated using the formula AU = Q - W. Plugging in the values we get, AU = (200 cal x 4.184 J/cal) - 1500 J = -630.64 J. Therefore, the internal energy of the water has decreased by 630.64 J.
b. To convert the change in internal energy into calories, we can use the formula 1 cal = 4.184 J. Plugging in the value we get, -630.64 J ÷ 4.184 J/cal = -150.74 cal. Therefore, the internal energy of the water has decreased by 150.74 cal.