What is the final temperature of each substance upon absorbing 2.25 kJ of heat?
You are almost correct, however, to calculate these values you need to put a bracket around he $t-29$ term (which is the $\Delta T$ term from $Q = mc \Delta T$). Hence, from your working out step:

$$2250 = 0.240 \times 28 \times (t - 29)$$

You'll need to expand the brackets.