Clouds form when water vapor out of the air

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

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

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

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

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

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

The graphs are shown below.

6 0

3 years ago

What phase of matter is least common on earth

Diano4ka-milaya [45]

Plasma is the least common matter on earth

4 0

3 years ago

What is abbreviated structural formula of the 2-metil-5-isopropiloctan?

Alja [10]

Answer:

CH3

|

CH3- C H -CH2-CH2- CH - CH2-CH2-CH3

|

CH

/ \

CH3 CH3

Explanation:

Octan

C-C-C-C-C-C-C-C

Metyl

CH3 -

Isopropyl

CH3

/

- CH

\

CH3

2-metil-5-isopropiloctan

CH3

|

CH3- C H -CH2-CH2- CH - CH2-CH2-CH3

|

CH

/ \

CH3 CH3

8 0

3 years ago

12.70 L of a gas has a pressure of 0.63 atm. What is the volume of the gas at 105kPa?

Zielflug [23.3K]

Answer: 7.693 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=0.63atm\\V_1=12.70L\\P_2=105kPa=1.04atm(1kPa=0.009atm)\\V_2=?$

Putting values in above equation, we get:

$0.63\times 12.70mL=1.04\times V_2\\\\V_2=7.693L$

Thus new volume of the gas is 7.693 L

6 0

3 years ago

Read 2 more answers

A mixture of gaseous reactants is put into a cylinder, where a chemical reaction tums them into gaseous products. The cylinder h

erastova [34]

Answer:

The reaction is exothermic

The temperature of the water bath goes up

Explanation:

An exothermic reaction is one in which energy flows out of the reaction system.

In this case, the system is the reaction vessel while the surrounding is the water bath. We see in the question that 300.1J of heat flows out of the system during the reaction. This is heat lost to the surroundings showing that the reaction is exothermic.

As energy is lost to the surroundings, the temperature of the water bath rises accordingly.

6 0

3 years ago