Question

An ice cube has a mass of 64 g and is initially at a temperature of 0°C . The ice cube is heated until 56.6 g has become water at 100°C and 7.4 g has become steam at 100°C. How much energy (in kJ) was transferred to the ice cube to accomplish the transformation?

Answer 1

The total energy transferred to the ice cube is calculated by summing the energies required for melting the ice, heating the water to 100°C, and vaporizing a portion of the water to steam at 100°C.

Step-by-step explanation:

The question refers to the amount of energy transfer required to transform an ice cube first into water and then partly into steam, covering phase changes and temperature changes. The energy needed for these changes can be calculated using specific heat capacities, heat of fusion, and heat of vaporization values.

Firstly, we calculate the energy needed to melt 56.6g of ice at 0°C to water at 0°C using the heat of fusion (which is 334 kJ/kg for water): Qmelting = mice x Lf, where mice is the mass of the ice and Lf is the latent heat of fusion.

Next, we calculate the energy needed to heat 56.6g of water from 0°C to 100°C using the specific heat capacity of water, which is 4.18 J/g°C: Qheating water = mwater x cwater x ΔT.

Finally, we calculate the energy needed to convert 7.4g of water at 100°C to steam at 100°C using the heat of vaporization (2256 kJ/kg for water): Qvaporization = msteam x Lv.

The sum of all three energy quantities will give us the total energy transferred.

Answer 2

The energy transferred to the ice cube to transform it into water at 100°C and steam at 100°C is 59.2 kJ. This is calculated by adding the energy required to melt the ice (18.9 kJ), heat the water to 100°C (23.6 kJ), and vaporize part of the water into steam (16.7 kJ).

Step-by-step explanation:

To calculate the amount of energy transferred to the ice cube during its transformation, we need to consider three distinct processes:

• Melting the ice to water at 0°C.
• Heating the resultant water from 0°C to 100°C.
• Vaporizing some of the water at 100°C to steam.

Let's break down the energy required for each step:

1. Melting of ice: Using Q = mLf, with m being the mass of ice at 0°C turned into water, and Lf being the latent heat of fusion, we get:
2. Q1 = (56.6 g)(334 J/g) = 18,904.4 J or 18.9 kJ (since 1 g of ice requires 334 J to melt).
3. Heating of melted water: The specific heat capacity (c) of water is approximately 4.18 J/g°C, so:
4. Q2 = (56.6 g)(4.18 J/g°C)(100°C) = 23,648.8 J or 23.6 kJ.
5. Vaporizing the water to steam: Using Q = mLv, with Lv being the latent heat of vaporization, we have:
6. Q3 = (7.4 g)(2260 J/g) = 16,724 J or 16.7 kJ (since 1 g of water requires 2260 J to vaporize).

Now, summing the total energy for all processes:

Total energy (Q total) = Q1 + Q2 + Q3

Total energy (Q total) = 18.9 kJ + 23.6 kJ + 16.7 kJ

Total energy (Q total) = 59.2 kJ

Therefore, the energy transferred to the ice cube to accomplish the transformation is 59.2 kJ.