Introduction:
In this problem, we are given 1kg of ice at -10 degrees Celsius and 1kg of water at 100 degrees Celsius. We need to determine the equilibrium temperature when the ice and water mix, as well as the content of the mixture. To solve this problem, we will use the principles of heat transfer and the concept of specific heat capacity.

Analysis:
When the ice and water are mixed, heat will flow from the water at 100 degrees Celsius to the ice at -10 degrees Celsius until thermal equilibrium is reached. This means that the final temperature of the mixture will lie somewhere between -10 and 100 degrees Celsius.

Calculating heat transfer:
To find the equilibrium temperature, we need to calculate the amount of heat transferred from the water to the ice. The amount of heat transferred can be calculated using the formula:

Q = mcΔT

where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Calculating heat transferred to the ice:
The ice will absorb heat until it reaches its melting point, 0 degrees Celsius. The amount of heat transferred to the ice can be calculated as follows:

Qice = m × cice × ΔTice

where Qice is the heat transferred to the ice, m is the mass of the ice, cice is the specific heat capacity of ice, and ΔTice is the change in temperature of the ice.

Calculating heat transferred to the water:
The water will lose heat until it reaches the equilibrium temperature. The amount of heat transferred to the water can be calculated using the same formula:

Qwater = m × cwater × ΔTwater

where Qwater is the heat transferred to the water, m is the mass of the water, cwater is the specific heat capacity of water, and ΔTwater is the change in temperature of the water.

Setting up the equation:
Since the total heat transferred to the ice and water must be equal, we can set up the equation:

Qice = Qwater

m × cice × ΔTice = m × cwater × ΔTwater

Solving the equation:
Substituting the given values, we have:

1kg × cice × (-10°C - 0°C) = 1kg × cwater × (100°C - 0°C)

Simplifying the equation gives:

cice × (-10°C) = cwater × 100°C

Dividing both sides by 100°C, we get:

cice × (-0.1) = cwater

Conclusion:
From the equation, we can see that the specific heat capacity of ice (cice) is equal to the specific heat capacity of water (cwater) multiplied by -0.1. Therefore, the equilibrium temperature will be 0 degrees Celsius, the melting point of ice. The content of the mixture will be a combination of melted ice and water.

Mixture content is 2 kg and 35degree celsius equilibrium temperature