A .6 gm peanut is burned beneath 50g of water, which increases the temperature from 22 degrees to 50 degrees centigrade. Assuming 40% efficiency, show that the peanut's food value is 3500 calories.
1 caloris is the amount of energy needed to raise the temp of 1g or water 1 degree.
So if it raises 50g by 28 deg, that's how many calories?
OK, so we know how many calories affected the water, but we are also told that the burning was only 40% efficient, which means that our answer is only 40% of the total energy in the peanut.
Is this enough?
A calorimetric measurement of food value in which the combustion is only 40% efficient does not sound very useful. How can one be sure of the efficiency?
jim has correctly shown you how to do the calculation and get 3500 (gram)-calories. You won't need the peanut's mass since they are not asking for calories per gram. Not that the units are not the Calories usually used for food.
1400
To calculate the food value of the peanut, we need to determine the amount of heat energy it provides when burned. We can use the equation:
Q = m * c * ΔT
where:
Q is the heat energy in calories,
m is the mass of the water in grams,
c is the specific heat capacity of water (1 calorie/gram·°C),
and ΔT is the change in temperature in °C.
First, let's calculate the mass of the water (m). We know that:
m = 50g
Next, we need to calculate ΔT:
ΔT = 50°C - 22°C = 28°C
Now, let's calculate Q using the formula above:
Q = 50g * 1 calorie/g·°C * 28°C = 1400 calories
This is the amount of heat energy transferred to the water. However, we need to take into account the efficiency of the peanut's food value, which is given as 40%.
To determine the food value of the peanut (FV), we divide the heat energy (Q) by the efficiency (E):
FV = Q / E
FV = 1400 calories / 0.4 = 3500 calories
Therefore, the food value of the peanut is 3500 calories.