What is the common temperature of the final equilibrium state of the system? A 10.0 kg block of ice at 0.0°C is placed into 10 kg water bath at 90.0C in an isolated container. The system eventually comes to equilibrium at a common temperature. Ignore the effect of the container.

The equilibrium temperature is 45°C

Explanation:

Given;

mass of block of ice, m₁ = 10 kg

mass of water, m₂ = 10 kg

initial temperature of the ice, t₁ = 0.0°C

initial temperature of water, t₂ = 90.0°C

let the equilibrium temperature = T

Apply the principles of mixture;

Heat lost by the water = heat gained by the ice

m₂c(90 - T) = m₁c(T - 0)

m₂(90 - T) = m₁(T - 0)

10(90 - T) = 10(T)

90 - T = T

90 = 2T

T = 90 / 2

T = 45°C

Therefore, the equilibrium temperature is 45°C

that would be because of both gravity and friction. so     c

because if the tool that you are using is not balanced it there would be false mass.