Athermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100of. after the thermometer reaches steady state, it is suddenly placed in a bath at 110of at t = 0 and left there for 1 min, after which it is immediately returned to the bath at 100of.(a) plot the variation of the thermometer reading with time (no hand-sketched plot ) calculate the thermometer reading at t = 0.5 min and at t = 2.0 min.

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

1. Calculate the thermometer readings after  0.5 min and 1 min

(a) After 0.5 min

(b) After 1 min

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

  t = t - 1, because the cooling starts 1 min late

3. Plot the temperature readings as a function of time.

The graphs are shown below.

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

it's filtration, but this is an assumption so my apologies if i got this