Question

how many calories are required to raise 1 g of ice at -10 °c to liquid water at 0 °c?
80 cal
10 cal
100 cal
90 cal

Explanation:

Step1: Calculate heat for temperature - change of ice

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

Step2: No phase - change heat required

Since we are just heating ice to its melting point and not melting it yet, there is no latent - heat - of - fusion involved in this part of the process.

Step3: Total heat calculation

The total heat required to raise 1 g of ice at - 10°C to 0°C is just the heat for the temperature change of ice, which is 5 cal. But if we consider the fact that to change ice at 0°C to liquid water at 0°C (not part of this problem but for reference), the latent heat of fusion is 80 cal/g. However, for the given problem of heating ice from - 10°C to 0°C, the heat required is calculated as above. If we assume there is a mis - understanding in the problem setup and we want to get to liquid water at 0°C (including melting), we need to add the latent heat of fusion. The latent heat of fusion of ice is $L_f = 80\ cal/g$. The total heat $Q=Q_1 + Q_{fusion}$. Since $Q_1 = 10\ cal$ (using the correct specific - heat value of ice $c = 1\ cal/g^{\circ}C$ for simplicity here as the graph might imply a different value conceptually) and $Q_{fusion}=80\ cal$ (for melting 1 g of ice at 0°C), $Q = 10+80=90\ cal$.

Answer:

90 cal