If all 1700 Joules of work get converted into heat, then, of course, you get 1700 Joules of heat.
The power needed can be calculated using the formula: Power = Energy / Time. Plugging in the given values, the power required to produce 1700 Joules in 5 seconds is 340 Watts.
The amount of heat energy produced from electrical energy depends on the efficiency of the system. If we assume 100% efficiency, then all 3000 joules of electrical energy would be converted into heat energy. Thus, 3000 joules of electrical energy would produce 3000 joules of heat energy.
Power is the rate at which work is done, and can be calculated as work divided by time. In this case, power = 100 Joules / 10 seconds = 10 watts. Therefore, 100 Joules of work generated 10 watts of power over a 10 second period.
The work done by the system can be calculated by finding the difference between the heat absorbed from the high-temperature reservoir and the heat passed onto the low-temperature reservoir. In this case, the work done by the system is 130 joules (425 joules - 295 joules).
You would need 20,920 Joules of heat to raise the temperature of 1kg of water by 5°C. This value is calculated using the specific heat capacity of water, which is 4186 J/kg°C.
The power needed can be calculated using the formula: Power = Energy / Time. Plugging in the given values, the power required to produce 1700 Joules in 5 seconds is 340 Watts.
The amount of heat energy produced from electrical energy depends on the efficiency of the system. If we assume 100% efficiency, then all 3000 joules of electrical energy would be converted into heat energy. Thus, 3000 joules of electrical energy would produce 3000 joules of heat energy.
It is measured in joules (J)
Something about 1550 foot pounds, or 2115 joules.
1000 watts is a measure of power, not heat. Power is the rate at which energy is used or transferred. To determine the amount of heat generated by 1000 watts, additional information about the time over which the power is used or the efficiency of the system is needed.
Power is the rate at which work is done, and can be calculated as work divided by time. In this case, power = 100 Joules / 10 seconds = 10 watts. Therefore, 100 Joules of work generated 10 watts of power over a 10 second period.
The necessary heat is 9,22 joules.
The latent heat of condensation of steam is 2260 Joules per gram (539.3 cals/g). So the amount of heat released by 12.4 g = 12.4*2260 Joules = 28,024 Joules or 6687 cals.
2.5 g 1 mol/18.02 g (-285.83) kJ/mol
The work done by the system can be calculated by finding the difference between the heat absorbed from the high-temperature reservoir and the heat passed onto the low-temperature reservoir. In this case, the work done by the system is 130 joules (425 joules - 295 joules).
q( in Joules ) = mass * specific heat * change in temperature [ convert temps--Tf = Tc(1.80) + 32 ] q = (40 g)(0.90 J/gC)(61.1o C - 22.8o C) = 1.4 X 103 Joules =============
The space shuttle required approximately 500 million Joules of energy for re-entry into the Earth's atmosphere. This energy was mainly used to slow down the shuttle and withstand the heat generated during re-entry.