The temperature of 1 mole of a liquid is raised by heating it with 750 joules of energy. It expands and does 200 joules of work, calculate the change in internal energy of the liquid.
I want to use the expression: $$\Delta U = \Delta Q + \Delta W$$ so that: $$\Delta U = 750\,\mathrm J- 200\,\mathrm J = 550\,\mathrm J$$
but it strikes me that it can't be that simple (first year college exam paper). Also, of what significance is the '1 mole' of liquid?
Your calculation is correct. The standardized definition of the change in internal energy $U$ for a closed thermodynamic system is
$$\Delta U=Q+W$$
where $Q$ is amount of heat transferred to the system and $W$ is work done on the system (provided that no chemical reactions occur). Therefore, heat transferred to the system is assigned a positive sign in the equation $$Q=750\ \mathrm J$$ whereas work done by the system on the surroundings during the expansion of the liquid is assigned a negative sign $$W=-200\ \mathrm J$$ Thus, the change in internal energy is $$\begin{align} \Delta U&=Q+W\\ &=750\ \mathrm J-200\ \mathrm J\\ &=550\ \mathrm J\\ \end{align}$$
However, the question is a bit flawed since the given values are not typical for a liquid. By way of comparison, realistic values for water are shown in the following table.
$$\textbf{Water (liquid)}\\ \begin{array}{lllll} \hline \text{Quantity} & \text{Symbol} & \text{Initial value (0)} & \text{Final value (1)} & \text{Change}\ (\Delta) \\ \hline \text{Amount of substance} & n & 1.00000\ \mathrm{mol} & 1.00000\ \mathrm{mol} & 0\\ \text{Volume} & V & 18.0476\ \mathrm{ml} & 18.0938\ \mathrm{ml} & 0.0462\ \mathrm{ml} \\ & & 1.80476\times10^{-5}\ \mathrm{m^3} & 1.80938\times10^{-5}\ \mathrm{m^3} & 4.62\times10^{-8}\ \mathrm{m^3} \\ \text{Pressure} & p & 1.00000\ \mathrm{bar} & 1.00000\ \mathrm{bar} & 0 \\ & & 100\,000\ \mathrm{Pa} & 100\,000\ \mathrm{Pa} & 0 \\ \text{Temperature} & T & 20.0000\ \mathrm{^\circ C} & 29.9560\ \mathrm{^\circ C} & 9.9560\ \mathrm{^\circ C} \\ & & 293.1500\ \mathrm{K} & 303.1060\ \mathrm{K} & 9.9560\ \mathrm{K} \\ \text{Internal energy} & U & 1\,511.59\ \mathrm{J} & 2\,261.58\ \mathrm{J} & 749.99\ \mathrm{J} \\ \text{Enthalpy} & H & 1\,513.39\ \mathrm{J} & 2\,263.39\ \mathrm{J} & 750.00\ \mathrm{J} \\ \hline \end{array}$$
When $1\ \mathrm{mol}$ of water with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting expansion is actually only $$\begin{align} \Delta V&=V_1-V_0\\ &=18.0938\ \mathrm{ml}-18.0476\ \mathrm{ml}\\ &=0.0462\ \mathrm{ml}\\ &=4.62\times10^{-8}\ \mathrm{m^3} \end{align}$$
The corresponding pressure-volume work is $$\begin{align} W&=p\Delta V\\ &=100\,000\ \mathrm{Pa}\times4.62\times10^{-8}\ \mathrm{m^3}\\ &=0.00462\ \mathrm J \end{align}$$ which is clearly below the value given in the question $(W=200\ \mathrm J)$.
The values given in the question are appropriate for a gas. For example, realistic values for nitrogen are shown in the following table.
$$\textbf{Nitrogen (gas)}\\ \begin{array}{lllll} \hline \text{Quantity} & \text{Symbol} & \text{Initial value (0)} & \text{Final value (1)} & \text{Change}\ (\Delta) \\ \hline \text{Amount of substance} & n & 1.00000\ \mathrm{mol} & 1.00000\ \mathrm{mol} & 0\\ \text{Volume} & V & 24.3681\ \mathrm{l} & 26.5104\ \mathrm{l} & 2.1423\ \mathrm{l} \\ & & 0.0243681\ \mathrm{m^3} & 0.0265104\ \mathrm{m^3} & 0.0021423\ \mathrm{m^3} \\ \text{Pressure} & p & 1.00000\ \mathrm{bar} & 1.00000\ \mathrm{bar} & 0 \\ & & 100\,000\ \mathrm{Pa} & 100\,000\ \mathrm{Pa} & 0 \\ \text{Temperature} & T & 20.0000\ \mathrm{^\circ C} & 45.7088\ \mathrm{^\circ C} & 25.7088\ \mathrm{^\circ C} \\ & & 293.1500\ \mathrm{K} & 318.8588\ \mathrm{K} & 25.7088\ \mathrm{K} \\ \text{Internal energy} & U & 6\,081.06\ \mathrm{J} & 6\,616.83\ \mathrm{J} & 535.77\ \mathrm{J} \\ \text{Enthalpy} & H & 8\,517.87\ \mathrm{J} & 9\,267.87\ \mathrm{J} & 750.00\ \mathrm{J} \\ \hline \end{array}$$
When $1\ \mathrm{mol}$ of nitrogen with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting pressure-volume work is
$$\begin{align} W&=p\Delta V\\ &=100\,000\ \mathrm{Pa}\times0.0021423\ \mathrm{m^3}\\ &=214.23\ \mathrm{J} \end{align}$$ The corresponding enthalpy balance $$\begin{align} \Delta H&=\Delta U+W\\ 750.00\ \mathrm{J}&=535.77\ \mathrm{J}+214.23\ \mathrm{J} \end{align}$$ is quite similar to the values of the question $(\Delta H=Q=750\ \mathrm J,$ $\Delta U=550\ \mathrm J,$ and $W=200\ \mathrm{J}).$
