Specific Heat and Enthalpy

Question

You put 10.0g ice into 22.0g liquid water at room temperature. The mixture at thermal equilibrium is made up of 75% liquid water and 25% ice. What was the temperature of the ice before mixing?

Answer 1

To find the initial temperature of the ice, we can use the principle of conservation of energy. The energy gained by the ice must be equal to the energy lost by the water. Let's first find the energy lost by the water, which is the same as the energy gained by the ice.

We know the specific heat capacity of water is 4.18 J/g°C, and the latent heat of fusion for ice is 334 J/g. Let's denote the mass of ice that has melted as m.

Energy lost by water = Energy gained by ice

Since 25% of the mixture remains ice, we know that 75% of the ice has melted.

m = 0.75 * 10.0g = 7.5g

Now, we can set up the energy equations:

For the water: Energy lost by water = (mass of water) * (specific heat capacity of water) * (change in temperature of water) Q_water = 22.0g * 4.18 J/g°C * (T_initial - 0°C)

For the ice: Energy gained by ice = (mass of melted ice) * (latent heat of fusion) + (mass of melted ice) * (specific heat capacity of water) * (change in temperature of melted ice) Q_ice = 7.5g * 334 J/g + 7.5g * 4.18 J/g°C * (0°C - T_initial)

Now, we can set these equations equal to each other:

22.0g * 4.18 J/g°C * (T_initial - 0°C) = 7.5g * 334 J/g + 7.5g * 4.18 J/g°C * (0°C - T_initial)

Solving for T_initial:

(22.0g * 4.18 J/g°C + 7.5g * 4.18 J/g°C) * T_initial = 7.5g * 334 J/g

T_initial = (7.5g * 334 J/g) / (22.0g * 4.18 J/g°C + 7.5g * 4.18 J/g°C)

T_initial ≈ -10.4°C

The initial temperature of the ice before mixing was approximately -10.4°C.