Almost unlimited. Pressure is force divided by area; if the heat is applied in a very small area (small volume), even a small amount of heat can cause a tremendous pressure.
The heat generated by 1700 joules of work depends on the efficiency of the process. In an ideal case where all the work is converted into heat, the heat generated would also be 1700 joules. However, in real-world scenarios, the heat generated would be less due to energy losses.
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)
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.
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 =============
To calculate the heat required to raise the temperature of 1 kilogram of air by 1 degree Celsius, you can use the specific heat capacity of air, which is approximately 1,005 joules per kilogram per degree Celsius (J/kg·°C). Therefore, the heat needed is about 1,005 joules. This value can vary slightly based on factors like humidity and pressure, but 1,005 J is a good estimate for dry air at room temperature.
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.
To calculate the work done by the system, we can use the formula for efficiency: Efficiency = Work output / Heat input. First, let's determine how much heat is not passed on to the lower temperature reservoir: 425 - 295 = 130 J. Thus, the work done by the system is the heat that is not passed on, which is 130 joules.
To calculate the heat absorbed by the water, you can use the formula: heat = mass * specific heat capacity * temperature change. First, determine the specific heat capacity of water (4.18 J/g°C). Then, plug in the values: heat = 15 g * 4.18 J/g°C * 3.0°C. The heat absorbed by the water is 188.1 Joules.
When water vapor condenses, it releases the latent heat of vaporization, which is around 2260 joules per gram. This heat energy warms the surrounding air as it is released during the condensation process.