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Heat transfer is the process of thermal energy spontaneously moving from a hotter object to a cooler one. There are three primary modes of heat transfer: conduction, convection, and radiation.
In the context of the problem, heat transfer occurs as we input energy into the krypton gas, increasing its temperature. When dealing with heat transfer, especially in thermodynamics problems, it's important to consider the surroundings and the system to determine how energy is distributed.
Here, the heat added ( Q ) is directly related to the temperature change and depends on the specific heat capacity and the mass of the gas. This is crucial for calculating the total heat transferred into the system, which in this example is 10710 calorics at constant volume and 17850 calorics at constant pressure.

Specific heat capacity (c) is a material-specific property that defines the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. Different substances will require different amounts of heat energy for the same change in temperature.
In the problem, krypton being monatomic has two distinct heat capacities: \(c_v\) for constant volume and \(c_p\) for constant pressure. These are important because they dictate how much energy is needed to achieve the desired temperature change of the gas. The heat added depends heavily on which heat capacity is in use: \(c_v\) is used at constant volume, and \(c_p\) is used at constant pressure. This distinction is essential as it affects the calculations of heat input and other associated thermodynamic properties.

The First Law of Thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. Instead, it can only change forms. This law can be expressed as \(\Delta U = Q - W\), where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added to the system, and \(W\) is the work done by the system.
In our exercise, this means calculating how energy added to the krypton gas through heat (Q) translates to changes in internal energy (\Delta U) and work done by the gas. At constant volume, the work done \(W\) is zero, meaning all the heat supplied goes into changing the internal energy. At constant pressure, part of the heat is used to do work in addition to changing the internal energy. The changes in internal energy for both cases end up being the same because they only depend on the temperature change.

In thermodynamics, the work done by or on a system is a measure of energy transfer when there is interaction with the surroundings that involves mechanical energy. The work done in a thermodynamic process can vary depending on the conditions such as constant volume or constant pressure.
For the exercise at constant volume, the work done is zero because the volume does not change, meaning no mechanical work is performed on or by the system. At constant pressure, however, there is a change in volume, and therefore work is done by the gas as it expands. This work can be calculated using \(W = Q - \Delta U\), leading to the result of 7140 calorics for constant pressure. Understanding work in these contexts helps clarify how energy distributions differ between different thermodynamic paths and the impact on a system's state.