Question

Calculate the quantity of heat required to raise the temperature of 1 g of ice from -10'C to110'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 0.001 kg of ice at -10°C to 0.001 kg of ice at 0°C:

Q_1=m_ic_i\Delta T= 0.001 kg\cdot2100\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot10^{\circ}C=21\ J.

Let's find the amount of heat required to convert 0.001 kg of ice at 0°C to 0.001 kg of water at 0°C:

$$Q_2=m_iL_f= 0.001 kg\cdot3.34\cdot10^5\ \dfrac{J}{kg}=334\ J.$$

Let's find the amount of heat required to convert 0.001 kg of water at 0°C to 0.001 kg of water at 100°C:

Q_3=m_wc_w\Delta T= 0.001 kg\cdot4180\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot100^{\circ}C=418\ J.

Let's find the amount of heat required to convert 0.001 kg of water at 100°C to 0.001 kg of steam at 100°C:

$$Q_4=m_wL_v= 0.001 kg\cdot2.264\cdot10^6\ \dfrac{J}{kg}=2264\ J.$$

Let's find the amount of heat required to convert 0.001 kg of steam at 100°C to 0.001 kg of steam at 110°C:

Q_5=m_sc_s\Delta T= 0.001 kg\cdot1996\ \dfrac{J}{kg\cdot \!^{\circ}C}\cdot10^{\circ}C=19.96\ J.

Finally, the total amount of the heat required to convert 0.001 kg of ice at -10°C to steam at 110°C:

$$Q=21\ J+334\ J+418\ J+2264\ J+19.96\ J=3056.96\ J.$$