Will Maynez burns a 0.6-g peanut beneath 50 g of water, which increases in temperature from 22°C to 50°C. (The specific heat capacity of water is 1.0 cal/g?°C.) a. Assuming that 40% of the heat released by the burn- ing peanut makes its way to the water (40% efficiency), show that the peanut’s food value is 3500 calories (equivalently, 3.5 Calories).

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Answer:

40% of  the energy release by the peanut is 3500 calories

Explanation:

One calorie is defined as the amount of energy required to increase the temperature of one gram of water for one degree Celsius (or one Kelvin)

Equation for energy gain by water is

Q = mcΔT

where, m is the mass of the object

c is the specific heat capacity

ΔT is the change in temperature

c =  1.0 cal/g?°C.

m = 50 g

ΔT = 50°C - 22°C

    = 28°C

Q = (50)× (1)× (28)

  = 1400calories

The peanut contain 1400calories of energy .

amount that 40% of energy is released to water ,

so,

Q = 1400 calories / 0.4

= 3500 calories

Therefore, 40% of  the energy release by the peanut is 3500 calories

The peanut’s food value is 3500 calories (equivalently, 3.5 Calories) is = 40% of  the energy release by the peanut is 3500 calories

What is Calorie?

When One calorie is refer as the amount of energy required to increase the temperature of one gram of water for one degree Celsius (or one Kelvin)

The Equation for energy gain by water is

That is Q = mcΔT

where, m is the mass of the object are

Then c is the specific heat capacity

After that ΔT is the change in temperature

c is = 1.0 cal/g?°C.

m is = 50 g

ΔT is = 50°C - 22°C

   After that, = 28°C

Q is = (50)× (1)× (28)

 That is = 1400calories

When The peanut contain 1400calories of energy.

Then The amount that 40% of energy is released to water,

so,

Q is = 1400 calories / 0.4

Then = 3500 calories

Thus, 40% of  the energy release by the peanut is 3500 calories

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