Answer :
a. A 4-inch thick piece of two-layer chocolate cake with frosting provides 2008 kJ of energy, which is equivalent to 479.68 Calories.
b. A diet of 2300 Calories per day is equivalent to 9612.2 kJ per day. To cool 26 g of sand from 98.2°C to 64.5°C, we need to calculate the heat using the specific heat capacity formula.
a. To convert 2008 kJ to Calories, we use the conversion factor: 1 kJ = 0.239005736 Cal. Therefore, 2008 kJ * 0.239005736 Cal/kJ ≈ 479.68 Calories. So, the cake provides approximately 479.68 Calories of energy.
b. To convert 2300 Calories to kilojoules, we use the conversion factor: 1 Calorie = 4.184 kJ. Therefore, 2300 Cal * 4.184 kJ/Cal ≈ 9612.2 kJ per day.
To calculate the heat required to cool 26 g of sand from 98.2°C to 64.5°C, we use the specific heat capacity formula:
Q = m * c * ΔT
where Q is the heat energy, m is the mass of the sand, c is the specific heat capacity of sand, and ΔT is the change in temperature.
The specific heat capacity of sand is typically around 0.2 J/g°C. So, we plug in the values:
Q = 26 g * 0.2 J/g°C * (98.2°C - 64.5°C) ≈ 26 g * 0.2 J/g°C * 33.7°C ≈ 174.04 J.
Therefore, approximately 174.04 Joules of heat energy are required to cool 26 g of sand from 98.2°C to 64.5°C.
learn more about temperature here:
brainly.com/question/7510619
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