Question

if you start with 1 g of ice at -10 °c, how many calories are needed to bring it to water vapor at 100 °c?
730 cal
820 cal
720 cal
800 cal

Explanation:

Step1: Heat ice to 0°C

The specific - heat of ice is 0.5 cal/g°C. The temperature change $\Delta T=0 - (- 10)=10^{\circ}C$. Using the formula $Q = mc\Delta T$ (where $m = 1g$, $c = 0.5$ cal/g°C), we have $Q_1=1\times0.5\times10 = 5$ cal.

Step2: Melt the ice

The latent - heat of fusion of ice is 80 cal/g. Using the formula $Q = mL_f$ (where $m = 1g$ and $L_f=80$ cal/g), we get $Q_2 = 1\times80=80$ cal.

Step3: Heat water from 0°C to 100°C

The specific - heat of water is 1 cal/g°C. The temperature change $\Delta T=100 - 0 = 100^{\circ}C$. Using the formula $Q = mc\Delta T$ (where $m = 1g$, $c = 1$ cal/g°C), we have $Q_3=1\times1\times100 = 100$ cal.

Step4: Vaporize the water

The latent - heat of vaporization of water is 540 cal/g. Using the formula $Q = mL_v$ (where $m = 1g$ and $L_v = 540$ cal/g), we get $Q_4=1\times540 = 540$ cal.

Step5: Calculate total heat

The total heat $Q=Q_1 + Q_2+Q_3+Q_4$. Substituting the values: $Q=5 + 80+100+540=725$ cal, which is closest to 720 cal.

Answer:

720 cal