Answer to Question #204331 in Physics for Morgan Hopkin

Question

Calculate the total amount of heat required to raise the temperature of 55 kg of ice from -16℃ to a vapor at 118℃.

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 55 kg of ice at -16°C to 55 kg of ice at 0°C:

$$Q_1=m_ic_i\Delta T= 55\ kg\cdot2100\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot16^{\circ}C=1848000\ J.$$

Let's find the amount of heat required to convert 55 kg of ice at 0°C to 55 kg of water at 0°C:

$$Q_2=m_iL_f= 55\ kg\cdot3.34\cdot10^5\ \dfrac{J}{kg}=18370000\ J.$$

Let's find the amount of heat required to convert 55 kg of water at 0°C to 55 kg of water at 100°C:

$$Q_3=m_wc_w\Delta T= 55\ kg\cdot4180\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot100^{\circ}C=22990000\ J.$$

Let's find the amount of heat required to convert 55 kg of water at 100°C to 55 kg of steam at 100°C:

$$Q_4=m_wL_v= 55\ kg\cdot2.264\cdot10^6\ \dfrac{J}{kg}=124520000\ J.$$

Let's find the amount of heat required to convert 55 kg of steam at 100°C to 55 kg of steam at 118°C:

$$Q_5=m_sc_s\Delta T= 55\ kg\cdot1996\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot18^{\circ}C=1976040\ J.$$

Finally, the total amount of the heat required to convert 55 kg of ice at -16°C to steam at 118°C:

$Q=1848000\ J+18370000\ J+22990000\ J+124520000\ J+1976040\ J,$

$$Q=1.69\cdot10^8\ J.$$