calculated the amount of heat required 5 gm of ice at 0 c to vapour at 100 c l ice=80 cal/gm,l steam =540 cal/gm s water=1 cal/gm/c

To determine the total amount of heat required to convert 5 grams of ice at 0°C to steam at 100°C, we need to consider several steps in the heating and phase change process. Each step involves a specific heat transfer, which we can calculate using the given values for latent heat and specific heat.

Steps to Calculate Total Heat Required

The process can be broken down into the following stages:

• Melting the ice to water at 0°C
• Heating the water from 0°C to 100°C
• Vaporizing the water to steam at 100°C

1. Melting the Ice

First, we need to melt the ice into water. The latent heat of fusion (L_ice) is given as 80 cal/gm. The heat required for this step can be calculated using the formula:

Q_melt = m × L_ice

Where:

• m = mass of ice = 5 gm
• L_ice = 80 cal/gm

Plugging in the values:

Q_melt = 5 gm × 80 cal/gm = 400 cal

2. Heating the Water

Next, we need to heat the resulting water from 0°C to 100°C. The specific heat of water (s_water) is 1 cal/gm/°C. The heat required for this step can be calculated using:

Q_heat = m × s_water × ΔT

Where:

• ΔT = change in temperature = 100°C - 0°C = 100°C

Substituting the values:

Q_heat = 5 gm × 1 cal/gm/°C × 100°C = 500 cal

3. Vaporizing the Water

Finally, we need to convert the water at 100°C to steam. The latent heat of vaporization (L_steam) is given as 540 cal/gm. The heat required for this step is:

Q_vaporize = m × L_steam

Using the values:

Q_vaporize = 5 gm × 540 cal/gm = 2700 cal

Calculating Total Heat Required

Now, we can sum up the heat required for all three stages:

Total Heat (Q_total) = Q_melt + Q_heat + Q_vaporize

Substituting the values we calculated:

Q_total = 400 cal + 500 cal + 2700 cal = 3600 cal

Final Result

The total amount of heat required to convert 5 grams of ice at 0°C to steam at 100°C is 3600 calories.