Question

500 g of water at a temperature of 73 °C is mixed with 45 g of steam at a temperature of 100 °C. The ratio of the initial mass of steam and the mass of steam at equilibrium of the mixture is

• A. 9 : 4
• B. 3 : 1
• C. 3 : 2
• D. 5 : 4

Hint:

In calorimetry problems involving phase changes (like condensation of steam), the key is to apply the principle of conservation of energy: "Heat lost by hot bodies = Heat gained by cold bodies." Remember that phase changes occur at a constant temperature (e.g., 100 °C for steam condensation), and the heat involved is given by $Q = mL_v$ (latent heat of vaporization) or $Q = mL_f$ (latent heat of fusion). For temperature changes, $Q = mc\Delta T$. The equilibrium temperature reveals if a phase change is complete or partial. If the equilibrium temperature is the boiling point of steam, it implies not all steam condensed.

Answer 1

The Correct Option is A

Step 1: Identify the given information and the goal.
Given: \begin{itemize} \item Mass of water ($m_w$) = 500 g \item Initial temperature of water ($T_w$) = 73 °C \item Mass of steam ($m_s$) = 45 g \item Initial temperature of steam ($T_s$) = 100 °C \item Equilibrium temperature ($T_{eq}$) = 100 °C (Since steam is present at equilibrium, the temperature must be 100 °C). \end{itemize} We need to find the ratio of the initial mass of steam to the mass of steam at equilibrium, i.e., $\frac{m_s}{m_{s,eq}}$. Step 2: State the necessary physical constants.
We will use the following standard values for specific heat and latent heat: \begin{itemize} \item Specific heat capacity of water ($c_w$) = 1 cal g$^{-1}$ °C$^{-1}$ \item Latent heat of vaporization of steam ($L_v$) = 540 cal g$^{-1}$ \end{itemize} Step 3: Calculate the heat gained by water to reach 100 °C.
The water at 73 °C will absorb heat to reach the equilibrium temperature of 100 °C.
$Q_{gained} = m_w c_w (T_{eq} - T_w)$ $Q_{gained} = 500 \text{ g} \times 1 \text{ cal g}^{-1} \text{ °C}^{-1} \times (100 \text{ °C} - 73 \text{ °C})$ $Q_{gained} = 500 \times 27 \text{ cal}$ $Q_{gained} = 13500 \text{ cal}$ Step 4: Calculate the mass of steam that condenses.
The heat gained by the water comes from the condensation of some mass of steam. Let $m_c$ be the mass of steam that condenses into water at 100 °C.
The heat lost by the condensing steam is:
$Q_{lost} = m_c L_v$ By the principle of calorimetry, Heat gained = Heat lost: $Q_{gained} = Q_{lost}$ $13500 \text{ cal} = m_c \times 540 \text{ cal g}^{-1}$ $m_c = \frac{13500}{540} \text{ g}$ $m_c = \frac{1350}{54} \text{ g}$ $m_c = 25 \text{ g}$ So, 25 g of steam condenses into water. Step 5: Calculate the mass of steam remaining at equilibrium.
The initial mass of steam was 45 g.
The mass of steam remaining at equilibrium ($m_{s,eq}$) is the initial mass minus the condensed mass:
$m_{s,eq} = m_s - m_c$ $m_{s,eq} = 45 \text{ g} - 25 \text{ g}$ $m_{s,eq} = 20 \text{ g}$ Step 6: Calculate the required ratio.
The ratio of the initial mass of steam and the mass of steam at equilibrium is: Ratio = $\frac{\text{Initial mass of steam}}{\text{Mass of steam at equilibrium}} = \frac{m_s}{m_{s,eq}}$ Ratio = $\frac{45 \text{ g}}{20 \text{ g}}$ Ratio = $\frac{9 \times 5}{4 \times 5}$ Ratio = $\frac{9}{4}$ The ratio is 9:4, which matches option (1). The final answer is $\boxed{\text{9 : 4}}$.