Calculate the energy given out by 200 g of iron cooling from 95°C to 35°C. Specific heat capacity of iron is 400 J/kg/K.

The mass of the iron is $m = 200 \, \text{g}$. Convert this to kilograms:

$$m = 200 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} = 0.2 \, \text{kg}$$

The specific heat capacity of iron is $c = 400 \, \text{J/kg/K}$.
The initial temperature is $T_1 = 95^\circ\text{C}$.
The final temperature is $T_2 = 35^\circ\text{C}$.

The change in temperature is the difference between the initial and final temperatures:

$$\Delta T = T_1 - T_2$$ $$\Delta T = 95^\circ\text{C} - 35^\circ\text{C} = 60^\circ\text{C}$$

Since a change of $1^\circ\text{C}$ is equal to a change of $1 \, \text{K}$, $\Delta T = 60 \, \text{K}$.

Substitute the values into the formula:

$$Q = (0.2 \, \text{kg}) \times (400 \, \text{J/kg/K}) \times (60 \, \text{K})$$ $$Q = 80 \, \text{J/K} \times 60 \, \text{K}$$ $$Q = 4800 \, \text{J}$$

The energy given out is $\text{4800 J}$.