The amount of heat (in cal)needed to bring 1 g of ice from T = -20 $\circ$ to 100 $\circ$ C and completely evaporate it into steam is

Question

Quizplus · Accepted Answer

To calculate the total heat required, we need to consider the heat needed to warm the ice from -20°C to 0°C, melt the ice into water, heat the water from 0°C to 100°C, and finally evaporate the water into steam. The specific heat of ice is approximately 0.5 cal/g°C, the heat of fusion of ice is 80 cal/g, the specific heat of water is 1 cal/g°C, and the heat of vaporization of water is 540 cal/g. 1. Heating ice from -20°C to 0°C: $Q_1 = m \cdot c_{ice} \cdot \Delta T = 1 \cdot 0.5 \cdot (0 - (-20)) = 10 \, \text{cal}$2. Melting ice at 0°C to water: $Q_2 = m \cdot L_{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}$4. Vaporizing water at 100°C: $Q_4 = m \cdot L_{vaporization} = 1 \cdot 540 = 540 \, \text{cal}$Total heat required: $Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 = 10 + 80 + 100 + 540 = 730 \, \text{cal}$

The Amount of Heat (In Cal)needed to Bring 1 G ∘\circ∘

The Amount of Heat (In Cal)needed to Bring 1 G $\circ$