In a process,$701 \, J$ of heat is absorbed by a system and $394 \, J$ of work is done by the system. What is the change in internal energy for the process (in $, J$)?

For an isochoric process,the first law of thermodynamics can be expressed as:

$A$ system is provided $50 \text{ J}$ of heat and work done on the system is $10 \text{ J}$. What is the change in internal energy (in $\text{ J}$)?

When $500 \ J$ of heat is supplied to a system at constant volume,the temperature of the system increases from $20^oC$ to $25^oC$. The change in internal energy of the system is ...... $J$.

$C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$. The enthalpy change $(\Delta H)$ for the above reaction at $27^{\circ}C$ is $-1366.5 \text{ kJ mol}^{-1}$. Calculate the internal energy change $(\Delta U)$ for the same reaction at this temperature.

Assertion : For an isothermal reversible process $Q = -W$,i.e.,work done by the system equals the heat absorbed by the system.
Reason : Enthalpy change $(\Delta H)$ is zero for an isothermal process.