500 g of water and 100 g of ice at 0°C are in a calorimeter whose water equivalent is 40 g. 10 g of steam at 100°C is added to it. Then water in the calorimeter is : (Latent heat of ice = 80 cal/g, Latent heat of steam = 540 cal/g)

We are given:

• Mass of water, $$ m_w = 500 $$ g at 0°C
• Mass of ice, $$ m_i = 100 $$ g at 0°C
• Water equivalent of calorimeter, $$ W = 40 $$ g (meaning the calorimeter behaves like 40 g of water for heat absorption)
• Mass of steam, $$ m_s = 10 $$ g at 100°C
• Latent heat of ice, $$ L_i = 80 $$ cal/g
• Latent heat of steam, $$ L_s = 540 $$ cal/g

We need to find the final mass of water in the calorimeter after adding steam. The specific heat capacity of water is $$ c = 1 $$ cal/g°C.

First, calculate the total heat released by the steam when it condenses and cools to 0°C. The steam undergoes two processes:

1. Condensation: Steam at 100°C condenses to water at 100°C, releasing latent heat. $$ Q_{\text{condense}} = m_s \times L_s = 10 \times 540 = 5400 \text{ cal} $$
2. Cooling: The condensed water cools from 100°C to 0°C, releasing sensible heat. $$ Q_{\text{cool}} = m_s \times c \times \Delta T = 10 \times 1 \times (100 - 0) = 1000 \text{ cal} $$

Total heat available from steam: $$ Q_{\text{available}} = Q_{\text{condense}} + Q_{\text{cool}} = 5400 + 1000 = 6400 \text{ cal} $$

Next, calculate the heat required to melt all the ice. The ice is at 0°C, and melting it to water at 0°C requires latent heat: $$ Q_{\text{melt all ice}} = m_i \times L_i = 100 \times 80 = 8000 \text{ cal} $$

Since $$ Q_{\text{available}} = 6400 $$ cal is less than $$ Q_{\text{melt all ice}} = 8000 $$ cal, the available heat is insufficient to melt all the ice. Therefore, only part of the ice melts, and the final temperature remains at 0°C.

The heat available (6400 cal) is used solely to melt ice because the calorimeter and the existing water are at 0°C, and the temperature does not change (so they absorb no heat). The mass of ice melted is: $$ \text{Mass of ice melted} = \frac{Q_{\text{available}}}{L_i} = \frac{6400}{80} = 80 \text{ g} $$

Thus, 80 g of ice melts into water, and the remaining ice is: $$ \text{Remaining ice} = 100 - 80 = 20 \text{ g} $$

Now, calculate the total water in the calorimeter:

• Original water: 500 g
• Water from melted ice: 80 g
• Water from condensed steam: 10 g (since the steam condenses and cools to 0°C)

Total water mass: $$ \text{Total water} = 500 + 80 + 10 = 590 \text{ g} $$

The calorimeter's water equivalent (40 g) does not add to the water mass; it only accounts for heat capacity. The remaining 20 g of ice is not water, so it is not included in the liquid water mass.

Hence, the water in the calorimeter is 590 g.

The options are: A. 580 g B. 590 g C. 600 g D. 610 g

So, the correct answer is Option B.