A mixture is formed by adding 0.1 kg of steam at 100°C to 2 kg of water at 20°C in a copper can (0.2 kg) also at 20°C. Determine the final temperature of the mixture.

Given:

• Mass of steam ($m_s$) = $0.1 \text{ kg}$
• Initial temperature of steam ($T_{s,i}$) = $100 \text{ °C}$
• Mass of water ($m_w$) = $2 \text{ kg}$
• Initial temperature of water ($T_{w,i}$) = $20 \text{ °C}$
• Mass of copper can ($m_c$) = $0.2 \text{ kg}$
• Specific heat capacity of copper ($c_c$) = $400 \text{ J/kgK}$
• Initial temperature of copper can ($T_{c,i}$) = $20 \text{ °C}$

Assumed constants:

• Specific heat capacity of water ($c_w$) = $4200 \text{ J/kgK}$
• Specific latent heat of vaporization of steam ($L_v$) = $2.26 \times 10^6 \text{ J/kg}$

Let the final temperature of the mixture be $T_f$.

The steam first condenses at $100 \text{ °C}$ and then the resulting water cools from $100 \text{ °C}$ to $T_f$.
Heat lost by condensation ($Q_1$) = $m_s L_v$
Heat lost by cooling of condensed water ($Q_2$) = $m_s c_w (100 - T_f)$
Total heat lost by steam ($Q_{lost}$):

$$Q_{lost} = Q_1 + Q_2$$ $$Q_{lost} = (0.1 \text{ kg} \times 2.26 \times 10^6 \text{ J/kg}) + (0.1 \text{ kg} \times 4200 \text{ J/kgK} \times (100 - T_f) \text{ K})$$ $$Q_{lost} = 226000 \text{ J} + (42000 - 420 T_f) \text{ J}$$ $$Q_{lost} = (268000 - 420 T_f) \text{ J}$$

The water and the copper can both heat up from $20 \text{ °C}$ to $T_f$.
Heat gained by water ($Q_3$) = $m_w c_w (T_f - 20)$
Heat gained by copper can ($Q_4$) = $m_c c_c (T_f - 20)$
Total heat gained ($Q_{gained}$):

$$Q_{gained} = Q_3 + Q_4$$ $$Q_{gained} = (2 \text{ kg} \times 4200 \text{ J/kgK} \times (T_f - 20) \text{ K}) + (0.2 \text{ kg} \times 400 \text{ J/kgK} \times (T_f - 20) \text{ K})$$ $$Q_{gained} = (8400 (T_f - 20)) \text{ J} + (80 (T_f - 20)) \text{ J}$$ $$Q_{gained} = (8400 T_f - 168000) \text{ J} + (80 T_f - 1600) \text{ J}$$ $$Q_{gained} = (8480 T_f - 169600) \text{ J}$$

According to the principle of calorimetry, heat lost equals heat gained:

$$Q_{lost} = Q_{gained}$$ $$268000 - 420 T_f = 8480 T_f - 169600$$

Rearrange the equation to solve for $T_f$:

$$268000 + 169600 = 8480 T_f + 420 T_f$$ $$437600 = 8900 T_f$$ $$T_f = \frac{437600}{8900}$$ $$T_f = 49.1685... \text{ °C}$$

Rounding to one decimal place:

$$T_f \approx \boxed{49.2 \text{ °C}}$$