Question

Find the amount of heat energy needed to convert 400 grams of ice at -38°C ... q will be the total heat required to turn ice at -38 C into steam at 160 C.

Answer 1

To convert 400 grams of ice at -38°C to steam at 160°C, the total heat required, q, is 1,284,480 J.

Step-by-step explanation:

To find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C, we need to consider the different phases and the changes in temperature involved.

1. First, we need to heat the ice from -38°C to its melting point, which is 0°C. The heat required for this is given by the formula Q₁ = m × C × ΔT, where m is the mass of the ice (in grams), C is the specific heat capacity of ice, and ΔT is the change in temperature. In this case, m = 400 g, C = 2.06 J/g°C, and ΔT = 0°C - (-38°C) = 38°C. Using the formula, we find Q₁ = 31,040 J.
2. Next, we need to melt the ice at 0°C. The heat required for this is given by the equation Q₂ = m × L_f, where L_f is the heat of fusion of ice. The heat of fusion of ice is 334 J/g. So, Q₂ = 400 g × 334 J/g = 133,600 J.
3. After the ice has melted, we need to heat the water from 0°C to 100°C. The heat required for this is given by the formula Q₃ = m × C × ΔT, where m is the mass of the water (in grams), C is the specific heat capacity of water, and ΔT is the change in temperature. In this case, m = 400 g, C = 4.184 J/g°C, and ΔT = 100°C - 0°C = 100°C. Using the formula, we find Q₃ = 167,360 J.
4. Next, we need to evaporate the water at 100°C. The heat required for this is given by the equation Q₄ = m × L_v, where L_v is the heat of vaporization of water. The heat of vaporization of water is 2260 J/g. So, Q₄ = 400 g × 2260 J/g = 904,000 J.
5. Finally, we need to heat the steam from 100°C to 160°C. The heat required for this is given by the formula Q₅ = m × C × ΔT, where m is the mass of the steam (in grams), C is the specific heat capacity of steam, and ΔT is the change in temperature. In this case, m = 400 g, C = 2.03 J/g°C, and ΔT = 160°C - 100°C = 60°C. Using the formula, we find Q₅ = 48,480 J.

To find the total heat required to convert the ice at -38°C to steam at 160°C, we add up the amounts of heat required for each step: Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅ = 31,040 J + 133,600 J + 167,360 J + 904,000 J + 48,480 J = 1,284,480 J.

Answer 2

To convert ice at -38°C to steam at 160°C, we need to consider the different stages of the phase transition and calculate the heat required for each stage.

Step-by-step explanation:

To determine the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C, we need to consider the different stages of the phase transition. First, we need to heat the ice from -38°C to 0°C. The heat required to raise the temperature of the ice can be calculated using the formula: Q = m * C * ΔT, where Q is the heat energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Next, we need to calculate the heat energy required to melt the ice at 0°C using the formula: Q = m * L, where L is the heat of fusion for water.

Finally, we need to calculate the heat energy required to raise the temperature of the water from 0°C to 160°C using the formula: Q = m * C * ΔT. By adding up these three stages, we can find the total amount of heat energy required.