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).

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

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j