How much energy is required to change a 40 g ice cube from ice at -25°C to steam at 105°C?

Answer in J.

To calculate the total energy required to change the ice cube from -25°C to 105°C, we need to consider the following steps:

1. Heating the ice from -25°C to 0°C: The energy required can be calculated using the formula Q = m * C * ΔT, where Q represents the energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature. For ice, the specific heat capacity is 2.09 J/g°C. So, the energy required for this step is Q1 = 40 g * 2.09 J/g°C * (0°C - (-25°C)).

2. Melting the ice at 0°C: The energy required to change the state of the matter from solid to liquid is given by Q = m * ΔHf, where ΔHf is the heat of fusion of ice. For water, the heat of fusion is 334 J/g. Thus, the energy required for this step is Q2 = 40 g * 334 J/g.

3. Heating the water from 0°C to 100°C: The energy required for this step can be calculated using Q = m * C * ΔT, where the specific heat capacity of water is 4.18 J/g°C. Therefore, the energy required is Q3 = 40 g * 4.18 J/g°C * (100°C - 0°C).

4. Evaporating the water at 100°C: The energy required to change the state of the matter from liquid to gas is given by Q = m * ΔHv, where ΔHv is the heat of vaporization of water. For water, the heat of vaporization is 2260 J/g. Hence, the energy required for this step is Q4 = 40 g * 2260 J/g.

5. Heating the steam from 100°C to 105°C: The energy required for this step can be calculated using Q = m * C * ΔT, where the specific heat capacity of steam is 2.03 J/g°C. So, the energy required is Q5 = 40 g * 2.03 J/g°C * (105°C - 100°C).

Now, we just need to add up the energies of each step to get the total energy required:

Total Energy = Q1 + Q2 + Q3 + Q4 + Q5

Substituting the values:

Total Energy = (40 g * 2.09 J/g°C * (0°C - (-25°C))) + (40 g * 334 J/g) + (40 g * 4.18 J/g°C * (100°C - 0°C)) + (40 g * 2260 J/g) + (40 g * 2.03 J/g°C * (105°C - 100°C))

After evaluating this expression, we will get the total energy required to change the 40 g ice cube from ice at -25°C to steam at 105°C.