Hot Box

At thermal equilibrium the copper block and air will settle at the same temperature - we need to start by finding this temperature.

The thermal energy change in each object will be $E = m S \Delta T$ , and these will be equal to each other since all the energy that leaves the copper block goes into the air. Here, $m$ is the mass, $S$ is the specific heat capacity and $\Delta T$ is the change in the temperature of the object.

The mass of the air is its density $\times$ volume = 1.2 $\text{kg}$ . So we have $1.2 \times 1 \times \Delta T_{\text{Air}} = 0.5 \times 0.4 \times \Delta T_{\text{Copper}}$ (using specific heat capacity in $\text{kJ/kg/K}$ and mass in $\text{kg}$ .

So the change in temperature of the copper is six times the change in temperature of the air. Conveniently, the combined change in temperature is 700 degrees, so we can see that the air temperature must rise by 100 degrees.

Now, as the volume is fixed, we have (using the ideal gas relationship) $\dfrac{\Delta P}{P} = \dfrac{\Delta T}{T}$ , therefore, the pressure increases by $\dfrac{100}{293} \times 100$ , (293 being the starting temperature in Kelvin). Evaluating this we find that the pressure rises by just 34%, and so the lid doesn't pop off.

Note I've neglected the volume taken up by the copper block in terms of the air it replaces, as this has a very small effect - we would have to reconsider this if the % rise was very close to 40%.