Question

How much heat is required to convert 20.0 [g] of ice at −10.0 ∘C to steam at 100.0 ∘C?

Answer 1

$$Q=Q_1+Q_2+Q_3+Q_4,$$ $$Q=m_ic_i\Delta T+m_iL_f+m_wc_w\Delta T+m_wL_v.$$

Let's find the amount of heat required to convert 0.02 kg of ice at -10°C to 0.02 kg of ice at 0°C:

Q_1=m_ic_i\Delta T=0.02\ kg\cdot2100\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot10^{\circ}C=420\ J.

Let's find the amount of heat required to convert 0.02 kg of ice at 0°C to 0.02 kg of water at 0°C:

$$Q_2=m_iL_f=0.02\ kg\cdot3.34\cdot10^5\ \dfrac{J}{kg}=6680\ J.$$

Let's find the amount of heat required to convert 0.02 kg of water at 0°C to 0.02 kg of water at 100°C:

Q_3=m_wc_w\Delta T=0.02\ kg\cdot4180\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot100^{\circ}C=8360\ J.

Let's find the amount of heat required to convert 0.02 kg of water at 100°C to 0.02 kg of steam at 100°C:

$$Q_4=m_wL_v=0.02\ kg\cdot2.264\cdot10^6\ \dfrac{J}{kg}=45280\ J.$$

Finally, the total amount of the heat required to convert 0.02 kg of ice at -10°C to steam at 100°C:

$$Q=420\ J+6680\ J+8360\ J+45280\ J=60740\ J.$$