7.0 kilograms of ice at 0 0C and are melted to water at 0 0C. Find the increase in internal energy of the ice as it is melted to water at 0 0C. Use 3.33x 105 J for Lf of ice. Express your answers in calories.
What is 7*Lf ?
Now for calories, multiply Joules by .239
To find the increase in internal energy of the ice as it melts to water, you can use the equation:
ΔE = mLf
where:
ΔE is the change in internal energy
m is the mass of the ice
Lf is the latent heat of fusion
In this case, the mass of the ice is given as 7.0 kilograms and the latent heat of fusion (Lf) is given as 3.33 × 10^5 J.
To convert the answer to calories, we need to use the conversion factor: 1 calorie = 4.184 J.
Let's calculate the increase in internal energy:
ΔE = (7.0 kg) × (3.33 × 10^5 J)
ΔE ≈ 2.331 × 10^6 J
Now, to convert from joules to calories, we use the conversion factor:
ΔE = (2.331 × 10^6 J) × (1 cal / 4.184 J)
ΔE ≈ 557,746 cal
Therefore, the increase in internal energy of the ice as it is melted to water at 0°C is approximately 557,746 calories.