The idle warmth of dissolving of water is 334 kJ/kg 

So 1 kg of water at 0 deg. C contains 334 kJ more warmth vitality than 1 kg of ice at 0 deg. C. 

The particular warmth (warm limit) of water varies with temperature yet not by much. In the temperature extend from 0 to 100 deg. C, it is approx. 4.2 kJ/(kg K). So 1 kg of water at 100 deg. C contains 4.2 x 100 = 420 kJ more warmth vitality than 1 kg of water at 0 deg. C. 

The inert warmth of dissipation of water is 2257 kJ/kg 

For 1 kg of steam at 100 deg. C contains 2257 kJ more warmth vitality than 1 kg of water at 100 deg. C. 

Therefore 1 kg of steam at 100 deg. C has 334 + 420 + 2257 = 3011 kJ more warmth vitality than 1 kg of ice at 0 deg. C. 

1 kg of ice and 1 kg of steam = 2 kg of blend. 3011 kJ in 2 kg = 3011/2 = 1505.5 kJ/kg. 

We require the temperature at which 1 kg of water (or ice or steam) has 1505.5 kJ more warmth vitality than 1 kg of ice at 0 deg. C. 

1505.5 - 334 = 1171.5 

So it is likewise the temperature at which 1 kg of water (or steam) has 1171.5 kJ more warmth vitality than 1 kg of water at 0 deg. C. 

1171.5/4.2 = 278.9 

278.9 greater than 100 

Hence the temperature of the mix is surely 100 deg. C. You require more vitality to bubble water than to get it from ice to breaking point!

Introduction:
The problem states that 1 kg of ice at -10°C is mixed with 1 kg of water at 100°C. We need to determine the equilibrium temperature and the composition of the mixture. To solve this problem, we will use the concept of heat transfer and the principle of energy conservation.

Step 1: Calculate the heat gained by the ice:
The ice needs to be heated from -10°C to its melting point at 0°C. The specific heat capacity of ice is 2.09 J/g°C. Since we have 1 kg of ice, the total heat gained by the ice can be calculated using the formula:

Q = m * c * ΔT

Where:
Q = Heat gained
m = Mass of the substance
c = Specific heat capacity
ΔT = Change in temperature

In this case, the change in temperature is 0°C - (-10°C) = 10°C. Plugging in the values, we get:

Q_ice = 1 kg * 2.09 J/g°C * 10°C = 20.9 kJ

Step 2: Calculate the heat lost by the water:
The water needs to be cooled from 100°C to its melting point at 0°C. The specific heat capacity of water is 4.18 J/g°C. Since we have 1 kg of water, the total heat lost by the water can be calculated using the same formula as before:

Q_water = 1 kg * 4.18 J/g°C * 100°C = 418 kJ

Step 3: Calculate the heat required for phase change:
To convert the ice at 0°C into water at 0°C, we need to provide heat for the phase change. The latent heat of fusion for ice is 334 J/g. Since we have 1 kg of ice, the total heat required for phase change can be calculated as:

Q_phase_change = 1 kg * 334 J/g = 334 kJ

Step 4: Calculate the total heat exchanged:
The total heat exchanged between the ice and water can be calculated by summing up the heat gained by the ice and the heat lost by the water:

Q_total = Q_ice + Q_water

Plugging in the values, we get:

Q_total = 20.9 kJ + 418 kJ = 438.9 kJ

Step 5: Calculate the equilibrium temperature:
The equilibrium temperature can be calculated using the principle of energy conservation. The total heat exchanged is equal to the heat required for phase change plus the heat gained by the ice after reaching its melting point, which can be expressed as:

Q_total = Q_phase_change + Q_ice_after_melting

Rearranging the equation, we can solve for Q_ice_after_melting:

Q_ice_after_melting = Q_total - Q_phase_change

Plugging in the values, we get:

Q_ice_after_melting = 438.9 kJ - 334 kJ = 104.9 kJ

Using the formula for heat gained, we can calculate the change in temperature for the ice after melting:

ΔT_ice_after_melting = Q_ice_after_melting / (m * c)

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