Answer to Question #161514 in Physics for Dk

Question

 How much energy is required to change a 40.0-g ice cube from ice at -10.0°C to steam at 110°C?

Answer 1

$$Q=Q_1+Q_2+Q_3+Q_4+Q_5,$$ $$Q=m_ic_i\Delta T+m_iL_f+m_wc_w\Delta T+m_wL_v+m_sc_s\Delta T.$$

Let's find the amount of heat required to convert 0.04 kg of ice at -10°C to 0.04 kg of ice at 0°C:

$$Q_1=m_ic_i\Delta T=0.04\ kg\cdot2100\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot10^{\circ}C=840\ J.$$

Let's find the amount of heat required to convert 0.04 kg of ice at 0°C to 0.04 kg of water at 0°C:

$$Q_2=m_iL_f=0.04\ kg\cdot3.34\cdot10^5\ \dfrac{J}{kg}=13360\ J.$$

Let's find the amount of heat required to convert 0.04 kg of water at 0°C to 0.04 kg of water at 100°C:

$$Q_3=m_wc_w\Delta T=0.04\ kg\cdot4180\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot100^{\circ}C=16720\ J.$$

Let's find the amount of heat required to convert 0.04 kg of water at 100°C to 0.04 kg of steam at 100°C:

$$Q_4=m_wL_v=0.04\ kg\cdot2.264\cdot10^6\ \dfrac{J}{kg}=90560\ J.$$

Let's find the amount of heat required to convert 0.04 kg of steam at 100°C to 0.04 kg of steam at 110°C:

$$Q_5=m_sc_s\Delta T=0.04\ kg\cdot1996\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot10^{\circ}C=798.4\ J.$$

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

$$Q=840\ J+13360\ J+16720\ J+90560\ J+798.4\ J=122278.4\ J.$$