Fig. 2.1 shows a melting ice cube of mass 0.3 kg placed in a container filled with warm water. Pure ice melts at a target temperature of 0 °C. Temperature of water drops from 44 °C to 20 °C. Specific heat capacity of water is 4200 J/kg°C. Latent heat of fusion of ice is 3.4 × 10⁵ J/kg.

The formula for latent heat of fusion is $Q_1 = m L_f$.
Given: mass of ice $m = 0.3 \text{ kg}$, latent heat of fusion $L_f = 3.4 \times 10^5 \text{ J/kg}$.

$$Q_1 = (0.3 \text{ kg}) \times (3.4 \times 10^5 \text{ J/kg})$$ $$Q_1 = 102000 \text{ J}$$

The formula for specific heat capacity is $Q_2 = mc\Delta T$.
Given: mass of water $m = 0.3 \text{ kg}$, specific heat capacity of water $c = 4200 \text{ J/kg}^\circ C$, temperature change $\Delta T = 20^\circ C - 0^\circ C = 20^\circ C$.

$$Q_2 = (0.3 \text{ kg}) \times (4200 \text{ J/kg}^\circ C) \times (20^\circ C)$$ $$Q_2 = 25200 \text{ J}$$

Total energy $Q_{total} = Q_1 + Q_2$.

$$Q_{total} = 102000 \text{ J} + 25200 \text{ J}$$ $$Q_{total} = 127200 \text{ J}$$

The total thermal energy absorbed by the ice is $\boxed{127200 \text{ J}}$.

d) To calculate the mass of water in which the ice was placed, we use the principle of conservation of energy: the heat lost by the warm water equals the heat gained by the ice (which is $Q_{total}$ from part c).

The formula for heat lost is $Q_{lost} = m_w c_w \Delta T_w$.
Given: specific heat capacity of water $c_w = 4200 \text{ J/kg}^\circ C$, temperature change of warm water $\Delta T_w = 44^\circ C - 20^\circ C = 24^\circ C$. Let $m_w$ be the mass of the warm water.

$$Q_{lost} = m_w \times (4200 \text{ J/kg}^\circ C) \times (24^\circ C)$$ $$Q_{lost} = m_w \times 100800 \text{ J/kg}$$

From part (c), the total heat gained by the ice is $Q_{total} = 127200 \text{ J}$.

$$Q_{lost} = Q_{total}$$ $$m_w \times 100800 \text{ J/kg} = 127200 \text{ J}$$

$$m_w = \frac{127200 \text{ J}}{100800 \text{ J/kg}}$$ $$m_w = 1.2619... \text{ kg}$$

Rounding to three significant figures:

$$m_w = 1.26 \text{ kg}$$

The mass of water in which the ice was placed is $\boxed{1.26 \text{ kg}}$.