A 2.622 g sample of almond is ignited and used to melt ice in a can.

a. if the almond sample melted 15.24g of ice,What is the heat produced in kcal/gram?
b. If the heat from the almond sample was used to heat 100.0 mL of water at 0.0 degree centigrade.What would be the final temperature of the water

To calculate the heat produced by the almond sample and the final temperature of the water, we need to use the concepts of heat transfer and specific heat.

a. To calculate the heat produced in kcal/gram, we can use the heat transfer equation:

Heat = mass * heat capacity * temperature change

First, we need to calculate the heat transfer from the almond sample to melt the ice. The heat capacity of ice is 2.06 kcal/g°C. The temperature change is the change in state from solid to liquid, which is 0°C. The mass of ice melted is given as 15.24g. Thus, the heat produced can be calculated as:

Heat = 15.24g * 2.06 kcal/g°C * 0°C
= 0 kcal

Hence, the heat produced by the almond sample is 0 kcal/gram.

b. To calculate the final temperature of the water after being heated by the almond sample, we can use the equation:

Heat gained by water = mass of water * specific heat * temperature change

The mass of water is given as 100.0 mL, but we need to convert it to grams. The density of water is approximately 1g/mL, so the mass of water is:

Mass of water = 100.0 mL * 1 g/mL
= 100.0 g

The specific heat capacity of water is 1.00 calorie/gram°C. The initial temperature of water is 0.0°C. The heat transfer from the almond sample is zero, as calculated in part a.

Heat gained by water = 100.0g * 1.00 calorie/g°C * temperature change

Since we want the final temperature, we can rearrange the equation:

Temperature change = Heat gained by water / (mass of water * specific heat)

Since the heat gain by the water is zero (as calculated in part a), the final temperature will remain at 0.0°C. Therefore, the final temperature of the water will be 0.0°C.