If the heat capacities of both ice and steam are 0.5 cal/g · Cº,the quantity of heat required to change 1 g of ice at -10ºC to steam at 120ºC is approximately
A)750 cal
B)735 cal
C)630 cal
D)620 cal
E)555 cal

Question

Quizplus · Accepted Answer

To calculate the total heat required, we need to consider the heat needed for each phase change and temperature change. The specific heat capacities of ice and steam are given as 0.5 cal/g·°C, and we need to know the heat of fusion (melting) of ice, which is 80 cal/g, and the heat of vaporization of water, which is 540 cal/g. The process involves heating ice from -10°C to 0°C, melting the ice, heating the water from 0°C to 100°C, and then vaporizing the water and heating the steam from 100°C to 120°C.1. Heating ice from -10°C to 0°C: $q_1 = m \cdot c_{ice} \cdot \Delta T = 1 \cdot 0.5 \cdot (0 - (-10)) = 5 \text{ cal}$2. Melting the ice: $q_2 = m \cdot \Delta H_{fusion} = 1 \cdot 80 = 80 \text{ cal}$3. Heating water from 0°C to 100°C: $q_3 = m \cdot c_{water} \cdot \Delta T = 1 \cdot 1 \cdot (100 - 0) = 100 \text{ cal}$ (Note: The specific heat capacity of water is 1 cal/g·°C, not given but commonly known.)4. Vaporizing the water: $q_4 = m \cdot \Delta H_{vaporization} = 1 \cdot 540 = 540 \text{ cal}$5. Heating steam from 100°C to 120°C: $q_5 = m \cdot c_{steam} \cdot \Delta T = 1 \cdot 0.5 \cdot (120 - 100) = 10 \text{ cal}$Adding all these together: $q_{total} = q_1 + q_2 + q_3 + q_4 + q_5 = 5 + 80 + 100 + 540 + 10 = 735 \text{ cal}$.

If the Heat Capacities of Both Ice and Steam Are