Answer to Question #160794 in Molecular Physics | Thermodynamics for Nataly

Question

What quantity of heat is needed to convert 2 kg of ice at -34°C to steam at 100°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 2 kg of ice at -34°C to 2 kg of ice at 0°C:

$$Q_1=m_ic_i\Delta T=2\ kg\cdot2100\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot34^{\circ}C=142800\ J.$$

Let's find the amount of heat required to convert 2 kg of ice at 0°C to 2 kg of water at 0°C:

$$Q_2=m_iL_f=2\ kg\cdot3.34\cdot10^5\ \dfrac{J}{kg}=668000\ J.$$

Let's find the amount of heat required to convert 2 kg of water at 0°C to 2 kg of water at 100°C:

$$Q_3=m_wc_w\Delta T=2\ kg\cdot4180\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot100^{\circ}C=836000\ J.$$

Let's find the amount of heat required to convert 2 kg of water at 100°C to 2 kg of steam at 100°C:

$$Q_4=m_wL_v=2\ kg\cdot2.264\cdot10^6\ \dfrac{J}{kg}=4528000\ J.$$

Finally, the total amount of the heat required to convert 2 kg of ice at -34°C to steam at 100°C:

$$Q=142800\ J+668000\ J+836000\ J+4528000\ J=6174800\ J.$$