Question

How much heat is required to change \(1.0 \mathrm{kg}\) of ice, originally at \(-20.0^{\circ} \mathrm{C},\) into steam at \(110.0^{\circ} \mathrm{C} ?\) Assume 1.0 atm of pressure.

Short answer

Answer: 3074.0 kJ

Step-by-step solution

Heating the ice from -20°C to 0°C

To find the heat needed in this step, we'll use the formula: \(Q = mc\Delta T\) - m: mass of the ice (1.0 kg) - c: specific heat capacity of ice (\(2.1 \frac{kJ}{kg\cdot °C}\)) - \(\Delta T\): temperature difference (-20°C to 0°C, so \(\Delta T = 20°C\)) \(Q_1 = (1.0kg)(2.1\frac{kJ}{kg\cdot °C})(20 °C) = 42.0 kJ\)

Melting the ice into water at 0°C

To find the heat needed in this step, use the formula: \(Q = mL_f\) - m: mass of the ice (1.0 kg) - \(L_f\): latent heat of fusion for water (\(334 \frac{kJ}{kg}\)) \(Q_2 = (1.0kg)(334\frac{kJ}{kg}) = 334.0 kJ\)

Heating the water from 0°C to 100°C

To find the heat needed in this step, use the formula: \(Q = mc\Delta T\) - m: mass of the water (1.0 kg) - c: specific heat capacity of water (\(4.18 \frac{kJ}{kg\cdot °C}\)) - \(\Delta T\): temperature difference (0°C to 100°C, so \(\Delta T = 100°C\)) \(Q_3 = (1.0kg)(4.18\frac{kJ}{kg\cdot °C})(100 °C) = 418.0 kJ\)

Turning the water to steam at 100°C

To find the heat needed in this step, use the formula: \(Q = mL_v\) - m: mass of the water (1.0 kg) - \(L_v\): latent heat of vaporization for water (\(2260 \frac{kJ}{kg}\)) \(Q_4 = (1.0kg)(2260\frac{kJ}{kg}) = 2260.0 kJ\)

Heating the steam from 100°C to 110°C

To find the heat needed in this step, use the formula: \(Q = mc\Delta T\) - m: mass of the steam (1.0 kg) - c: specific heat capacity of steam (\(2.0 \frac{kJ}{kg\cdot °C}\)) - \(\Delta T\): temperature difference (100°C to 110°C, so \(\Delta T = 10°C\)) \(Q_5 = (1.0kg)(2.0\frac{kJ}{kg\cdot °C})(10 °C) = 20.0 kJ\)

Summing the heat involved in all phases

Now we'll add the heat involved in all the steps to get the total heat required. \(Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 42.0 kJ + 334.0 kJ + 418.0 kJ + 2260.0 kJ + 20.0 kJ = 3074.0 kJ\) So, the amount of heat required to change 1.0 kg of ice originally at -20°C into steam at 110°C is 3074.0 kJ.