Question

will maynez burns a 0.6 - g peanut beneath 58 g of water, which increases in temperature from 22°c to 46°c. (the specific heat capacity of water is 1.0 cal/g - °c.) part a assuming 32 % efficiency, find the peanuts food value. express your answer to two significant figures and include the appropriate units. part b then find the food value in calories per gram. express your answer using two significant figures.

Explanation:

Step1: Calculate heat absorbed by water

Use the formula $Q_{abs}=mc\Delta T$, where $m = 58\ g$, $c=1.0\ cal/g - ^{\circ}C$, and $\Delta T=(46 - 22)^{\circ}C$.
$Q_{abs}=58\ g\times1.0\ cal/g - ^{\circ}C\times(46 - 22)^{\circ}C=58\times24\ cal = 1392\ cal$

Step2: Calculate the food - value of the peanut

Since the efficiency $\eta = 32\%=0.32$, and $\eta=\frac{Q_{abs}}{Q_{peanut}}$. Then $Q_{peanut}=\frac{Q_{abs}}{\eta}$.
$Q_{peanut}=\frac{1392\ cal}{0.32}=4350\ cal\approx4.4\times 10^{3}\ cal$ (to two significant figures)

Step3: Calculate food - value per gram

The mass of the peanut is $m_{peanut}=0.6\ g$. The food - value per gram $q=\frac{Q_{peanut}}{m_{peanut}}$.
$q=\frac{4350\ cal}{0.6\ g}=7250\ cal/g\approx7.3\times 10^{3}\ cal/g$ (to two significant figures)

Answer:

Part A: $4.4\times 10^{3}\ cal$
Part B: $7.3\times 10^{3}\ cal/g$