Question

Five ice cubes, each with a mass of \(22.0 \mathrm{g}\) and at a temperature of \(-50.0^{\circ} \mathrm{C},\) are placed in an insulating container. How much heat will it take to change the ice cubes completely into steam?

Short answer

Answer: 3.825 × 10^5 J

Step-by-step solution

Calculate the heat to raise the temperature of the ice to \(0^{\circ}\mathrm{C}\).

First, we need to heat the ice from \(-50.0^{\circ}\mathrm{C}\) to \(0^{\circ}\mathrm{C}\). We can calculate the amount of heat needed using the formula: \(Q = mc\Delta{T}\), where \(Q\) is the heat, \(m\) is the mass, \(c\) is the specific heat capacity and \(\Delta{T}\) is the change in temperature. For ice, the specific heat capacity is \(c_{ice} = 2100 \mathrm{J/(kg \cdot K)}\). The mass of five ice cubes is \(m = 5 \times 22.0 \mathrm{g} = 110 \mathrm{g} = 0.110 \mathrm{kg}\). The temperature change is \(\Delta{T} = 0^{\circ}\mathrm{C} - (-50.0^{\circ}\mathrm{C}) = 50.0\mathrm{K}\). Now, we can calculate the heat needed: \(Q_1 = (0.110\,\mathrm{kg})(2100\,\mathrm{J/(kg \cdot K)})(50.0\,\mathrm{K}) = 1.155 \times 10^4\,\mathrm{J}\).

Calculate the heat to melt the ice at \(0^{\circ}\mathrm{C}\).

Next, we need to melt the ice at \(0^{\circ}\mathrm{C}\) to water using the formula for latent heat: \(Q = mL\), where \(Q\) is the heat, \(m\) is the mass, and \(L\) is the latent heat of fusion. For ice, the latent heat of fusion is \(L_{f} = 334 \times 10^3 \mathrm{J/kg}\). The mass of the ice cubes remains the same, \(m = 0.110\,\mathrm{kg}\). Now, calculate the heat needed to melt the ice: \(Q_2 = (0.110\,\mathrm{kg})(334 \times 10^3\,\mathrm{J/kg}) = 3.674 \times 10^4\,\mathrm{J}\).

Calculate the heat to heat the water from \(0^{\circ}\mathrm{C}\) to \(100^{\circ}\mathrm{C}\) and convert it to steam.

First, we need to heat up the water from \(0^{\circ}\mathrm{C}\) to \(100^{\circ}\mathrm{C}\). For water, the specific heat capacity is \(c_{water} = 4190\,\mathrm{J/(kg \cdot K)}\). The temperature change is \(\Delta{T} = 100^{\circ}\mathrm{C} - 0^{\circ}\mathrm{C} = 100\,\mathrm{K}\). Calculate the heat to heat the water: \(Q_3 = (0.110\,\mathrm{kg})(4190\,\mathrm{J/(kg \cdot K)})(100\,\mathrm{K}) = 4.603 \times 10^4\,\mathrm{J}\). Now, we need to convert the water at \(100^{\circ}\mathrm{C}\) to steam. The latent heat of vaporization for water is \(L_{v} = 2.26 \times 10^6\,\mathrm{J/kg}\). The mass of the water remains the same, \(m = 0.110\,\mathrm{kg}\). Calculate the heat needed to convert the water into steam: \(Q_4 = (0.110\,\mathrm{kg})(2.26 \times 10^6\,\mathrm{J/kg}) = 2.486 \times 10^5\,\mathrm{J}\).

Find the total heat needed.

Finally, we can find the total heat needed by adding the heat values from Steps 1-4: \(Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 = 1.155 \times 10^4\,\mathrm{J} + 3.674 \times 10^4\,\mathrm{J} + 4.603 \times 10^4\,\mathrm{J} + 2.486 \times 10^5\,\mathrm{J} = 3.825 \times 10^5\,\mathrm{J}\) It will take \(3.825 \times 10^5\,\mathrm{J}\) of heat to change the ice cubes completely into steam.