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(C) According to the First Law of Thermodynamics, $\Delta Q = \Delta U + \Delta W$, where $\Delta U$ is the change in internal energy.Since internal e

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$160 \ J$

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(C) According to the First Law of Thermodynamics, $\Delta Q = \Delta U + \Delta W$, where $\Delta U$ is the change in internal energy.Since internal energy is a state function, $\Delta U$ is the same for any path between the same two states $A$ and $C$.For the direct path $A \rightarrow C$:$\Delta W_{AC} = 200 \ J$$\Delta Q_{AC} = 280 \ J$$\Delta U = \Delta Q_{AC} - \Delta W_{AC} = 280 \ J - 200 \ J = 80 \ J$.For the path $A \rightarrow B \rightarrow C$:$\Delta W_{ABC} = 80 \ J$Since $\Delta U$ is a state function, $\Delta U = 80 \ J$ for this path as well.Therefore, the heat absorbed $\Delta Q_{ABC} = \Delta U + \Delta W_{ABC} = 80 \ J + 80 \ J = 160 \ J$.

Similar Questions

Write the sign convention for internal energy in thermodynamics.

In a thermodynamic process,$200 \ J$ of heat is given to a gas and $100 \ J$ of work is also done on it. The change in internal energy of the gas is ........ $J$.

The internal energy change in a system that has absorbed $2 \; kcal$ of heat and done $500 \; J$ of work is ...... $J$.

If $\Delta Q$ and $\Delta W$ represent the heat supplied to the system and the work done on the system respectively,then the first law of thermodynamics can be written as,where $\Delta U$ is the change in internal energy.

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