A 500 g aluminum calorimeter contains 300 g of liquid X. An electric heater with a current of 3.0 A and voltage of 220 V is used to heat the liquid for 2 minutes. If the initial temperature of the liq

Mass of aluminum calorimeter, $m_{cal} = 500 \text{ g} = 0.5 \text{ kg}$
Specific heat capacity of aluminum, $c_{cal} = 900 \text{ J/kg K}$
Mass of liquid X, $m_X = 300 \text{ g} = 0.3 \text{ kg}$
Current, $I = 3.0 \text{ A}$
Voltage, $V = 220 \text{ V}$
Time, $t = 2 \text{ minutes} = 2 \times 60 \text{ s} = 120 \text{ s}$
Initial temperature of liquid X and calorimeter, $T_{initial}$
Final temperature of liquid X and calorimeter, $T_{final} = 80^\circ \text{C}$

The heat supplied by the heater is given by the formula $Q = VIt$.

The heat gained by the calorimeter is given by $Q_{cal} = m_{cal} c_{cal} \Delta T$.
The temperature rise for the calorimeter is $\Delta T = T_{final} - T_{initial} = (80 - T_{initial})^\circ \text{C}$.

The heat gained by liquid X is given by $Q_X = m_X c_X \Delta T$.
The specific heat capacity of liquid X is $c_X = 2800 \text{ J/kg K}$.
The temperature rise for liquid X is $\Delta T = T_{final} - T_{initial} = (80 - T_{initial})^\circ \text{C}$.

The heat supplied by the heater is equal to the total heat gained by the calorimeter and liquid X.

Rounding to one decimal place:

The initial temperature of the liquid is $\text{18.6}^\circ \text{C}$.