150g of ice at 0 degreecelcius is mixed with 300g of water at 50 degreecelcius.calculate the temperature of the mixture.

the sum of heats gained is zero.

Hf(250)+150*cw*(Tf-0)+300*cw*(Tf-50)=0
put the heat of fusion in, and solve for Tf

How?

To calculate the temperature of the mixture, you can use the principle of conservation of energy. The total heat gained by the ice is equal to the total heat lost by the water.

First, let's calculate the heat gained by the ice using the formula:

Q = m × c × ΔT

Where:
Q is the heat gained or lost
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

For ice, the specific heat capacity is 2.09 J/g°C, and the change in temperature is the final temperature (Tm) minus the initial temperature (0°C).

Q_ice = 150g × 2.09 J/g°C × (Tm - 0°C)

Next, let's calculate the heat lost by the water:

Q_water = 300g × 4.18 J/g°C × (Tm - 50°C)

For water, the specific heat capacity is 4.18 J/g°C.

Since the total heat gained by the ice equals the total heat lost by the water, we can set the two equations equal to each other and solve for Tm:

150g × 2.09 J/g°C × (Tm - 0°C) = 300g × 4.18 J/g°C × (Tm - 50°C)

Solving this equation will give us the temperature of the mixture.