No, HEAT is work done!
When work is done on a system by pushing it, the internal energy of the system increases, leading to an increase in temperature. The work done increases the kinetic energy of the particles in the system, causing them to move faster and leading to an increase in temperature.
When work is done by a system with no heat added, the temperature of the system generally decreases. This is due to the fact that work done by the system often involves the system losing energy in the form of work, causing its internal energy and therefore its temperature to decrease.
In an isothermal process, the temperature remains constant, so work is done slowly to maintain this temperature. In an adiabatic process, there is no heat exchange with the surroundings, so work is done quickly, causing a change in temperature.
In an isothermal process, the temperature of the system remains constant. Since work done is the result of a change in energy, and the temperature does not change, there is no transfer of energy in the form of work during an isothermal process. Thus, the work done in an isothermal system is zero.
remains constant From Rafaelrz. When a simple closed system does work and no heat is added, the temperature of the system will drop. This is because the work is done at the expense of his internal energy, which is thermal energy.
When work is done on a system by pushing it, the internal energy of the system increases, leading to an increase in temperature. The work done increases the kinetic energy of the particles in the system, causing them to move faster and leading to an increase in temperature.
When work is done by a system with no heat added, the temperature of the system generally decreases. This is due to the fact that work done by the system often involves the system losing energy in the form of work, causing its internal energy and therefore its temperature to decrease.
When a system is doing work, it can either increase or decrease in temperature depending on the type of work being done. If work is done on the system, its temperature may increase due to the input of energy. Conversely, if the system is doing work on its surroundings, it may lose energy and decrease in temperature.
In an isothermal process, the temperature remains constant, so work is done slowly to maintain this temperature. In an adiabatic process, there is no heat exchange with the surroundings, so work is done quickly, causing a change in temperature.
In an isothermal process, the temperature of the system remains constant. Since work done is the result of a change in energy, and the temperature does not change, there is no transfer of energy in the form of work during an isothermal process. Thus, the work done in an isothermal system is zero.
remains constant From Rafaelrz. When a simple closed system does work and no heat is added, the temperature of the system will drop. This is because the work is done at the expense of his internal energy, which is thermal energy.
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).
In an isothermal process, the work done is the product of the pressure and the change in volume of the system. This is because the temperature remains constant throughout the process, so the work done is solely determined by the change in volume.
If a reaction is carried out in constant temperature and constant volume no work is done and heat exchanged with surroundings is equal to the internal energy.
When work is done on a substance, its internal energy increases. This can manifest as an increase in temperature, changes in phase, or changes in chemical composition. The work done on the substance can be in the form of mechanical work, electrical work, or other forms of energy input.
An increase in temperature does not require work to be done because it is a spontaneous process driven by the thermal energy of the system. Work is only required if we want to decrease the temperature or maintain a lower temperature against a natural increase.
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.