Final answer:
To convert ice to steam using a given amount of heat, we need to consider the heat required for each phase change. By calculating the heat needed to melt the ice and the heat needed to vaporize the water, we can determine the total heat required to convert the ice to steam. Using the given heat of 8,195 cal, we find that 13.2 grams of ice can be converted to steam.
Step-by-step explanation:
To calculate the number of grams of ice that can be converted to steam using a given amount of heat, we need to consider the heat required for each phase change. First, we calculate the heat needed to melt the ice at 0 °C by multiplying the mass of ice by the heat of fusion of water (79.9 cal/g). Then, we calculate the heat needed to vaporize the melted water by multiplying the mass of water by the heat of vaporization of water (540 cal/g). Finally, we add the two heats together to get the total heat required to convert the ice to steam. Let's assume the mass of ice is x grams. The heat required to melt the ice is x * 79.9 cal. After melting, the mass of water is also x grams. The heat required to vaporize the water is x * 540 cal. The total heat required is x * 79.9 cal + x * 540 cal. We can set this equal to the given heat of 8,195 cal and solve for x: x * 79.9 cal + x * 540 cal = 8,195 cal. Simplifying the equation: x * (79.9 cal + 540 cal) = 8,195 cal, x * 619.9 cal = 8,195 cal. Dividing both sides by 619.9 cal: x = 13.2 g. Therefore, 13.2 grams of ice at 0 °C can be converted to steam at 100 °C using 8,195 cal of heat.