Dry saturated steam at 100 degree C is passed into 250 gm of a mixture of ice and water contained in a calorimeter of thermal capacity 45J/K.When all the ice has just melted,the mass of the content has increased by 10 gm due to condensed steam.Assuming no heat exchange between the calorimeter and the surroundings,calculate

(a) the mass of ice initially in the calorimeter,and
(b) the rise in temperature when the passage of steam is continued until a further 15 gm of steam has condensed and the mixture is in thermal equilibrium.Specific latent heat of vaporization of water is 2260000 J/kg and specific heat capacity of water is 4200J/kg.K.

(a) Mass of ice initially in the calorimeter = 250 g - 10 g = 240 g

(b) Rise in temperature = (45 J/K x (25 g + 15 g)) / (4200 J/kg.K x (250 g + 25 g))
= 0.068 K

To solve this problem, we can use the principle of energy conservation.

(a) First, let's determine the heat absorbed by the ice initially in the calorimeter.

The heat absorbed by the ice can be calculated using the equation:

Q_ice = m_ice * latent heat of fusion

where Q_ice is the heat absorbed by the ice, m_ice is the mass of ice initially in the calorimeter, and latent heat of fusion is the specific latent heat of fusion of ice (334,000 J/kg).

Q_ice = m_ice * latent heat of fusion

Since the mass of the mixture increased by 10 grams due to condensed steam, we can write:

m_water = 250 grams - 10 grams

Now, we know that the heat absorbed by the mixture (water + ice) is equal to the heat provided by the steam:

Q_mixture = Q_steam

The heat absorbed by the mixture can be calculated using the equation:

Q_mixture = m_mixture * specific heat capacity of water * ΔT

where Q_mixture is the heat absorbed by the mixture, m_mixture is the mass of the mixture, specific heat capacity of water is 4200 J/kg.K, and ΔT is the change in temperature of the mixture.

ΔT = (Q_mixture) / (m_mixture * specific heat capacity of water)

Now, we can equate both equations:

m_ice * latent heat of fusion = m_mixture * specific heat capacity of water * ΔT

Substituting the known values, we have:

m_ice * 334,000 J/kg = (250 g - 10 g) * 4200 J/kg.K * ΔT

Now, let's solve for m_ice:

m_ice = [(250 g - 10 g) * 4200 J/kg.K * ΔT] / (334,000 J/kg)

(b) To calculate the rise in temperature when the passage of steam is continued until a further 15 g of steam has condensed and the mixture is in thermal equilibrium, we use the same equation:

ΔT = (Q_mixture) / (m_mixture * specific heat capacity of water)

The heat absorbed by the mixture is now the heat provided by the steam and can be calculated as:

Q_mixture = m_mixture * specific heat capacity of water * ΔT + m_steam * latent heat of vaporization

where m_steam is the mass of the condensed steam (15 g) and latent heat of vaporization is the specific latent heat of vaporization of water (2,260,000 J/kg).

ΔT = (m_mixture * specific heat capacity of water * ΔT + m_steam * latent heat of vaporization) / (m_mixture * specific heat capacity of water)

Now, let's solve for ΔT:

ΔT = [(m_steam * latent heat of vaporization) / (m_mixture * specific heat capacity of water - m_steam * specific heat capacity of water)]

Substituting the known values, we have:

ΔT = [(15 g * 2,260,000 J/kg) / ((250 g - 10 g) * 4200 J/kg.K - 15 g * 4200 J/kg.K)]

Note: Make sure to convert all mass values to kilograms and temperature values to Kelvin to maintain consistent units throughout the calculations.

To solve this question, we will use the principles of heat transfer and phase change.

(a) To determine the initial mass of ice, we need to find the heat gained by the ice to melt it. This can be calculated using the equation:

Q = mL

where Q is the heat gained, m is the mass, and L is the specific latent heat of fusion (melting) of water.

We know that the calorimeter's thermal capacity is 45 J/K. Therefore, the heat gained by the calorimeter during this process can be written as:

Q_calorimeter = CΔT

where C is the thermal capacity of the calorimeter and ΔT is the change in temperature.

During the process, all of the ice will melt and raise the temperature of the water from 0°C to 100°C. Therefore, we have:

Q_calorimeter = (m_ice + m_water) * c_water * ΔT_water

where m_ice is the initial mass of ice, m_water is the mass of water initially present in the calorimeter, c_water is the specific heat capacity of water, and ΔT_water is the change in temperature of the water.

At thermal equilibrium, the heat gained by the condensed steam will be equal to the heat lost by the steam to reach its saturation temperature (100°C). We can write this as:

Q_condensed steam = m_condensed steam * L_vaporization

where m_condensed steam is the mass of the condensed steam and L_vaporization is the specific latent heat of vaporization of water.

From the question, we know that the mass of the content has increased by 10g due to the condensed steam. Therefore, we have:

Q_calorimeter + Q_condensed steam = 0

Substituting the values and equations, we get:

(m_ice + m_water) * c_water * ΔT_water + m_condensed steam * L_vaporization = 0

Substituting the given values, which are:
m_water = 250g
C = 45 J/K
ΔT_water = 100°C
L_vaporization = 2260000 J/kg

We solve for m_ice:

(250g + m_ice) * 4200 J/kg.K * (100 - 0)°C + 10g * 2260000 J/kg = 0

Solve the equation to find the value of m_ice.

(b) To find the rise in temperature when a further 15g of steam has condensed and the mixture is in thermal equilibrium, we will use the law of conservation of energy. The heat gained by the condensed steam should be equal to the heat lost by the steam and gained by the mixture.

Q_condensed steam = Q_mixture

Substituting the values, we get:

m_condensed steam * L_vaporization = (m_ice + m_water) * c_water * ΔT_mixture

Solving this equation for ΔT_mixture will provide the rise in temperature.

You can now calculate the answers by substituting the given values into the equations.