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Trava [24]
3 years ago
8

To warm 2.0 l of tea (d = 1.01 g/ml; sp. heat = a cook places a 500 g block of stone at a temperature of 200f into the teapot. a

ssuming that the tea was initially at 72f, what is the final temperature of the tea in f?
a. 33
b. 91
c. 110
d. 140
e. 200
Physics
1 answer:
IRISSAK [1]3 years ago
4 0
Volume of tea V = 2.0L = 2000 mL density d = 1.01 g/ mL mass of tea m = V * d = 2000mL * 1.01g/mL = 2020 gWhen we assume that the tea was initially at 72, the final temperature of the tea in F is 91. 
The answer in this question is B. 91
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Now we know that the length of each side is 2 and we can find the area of one side by doing 2x2 which is 4.

To find the total surface area of one cube we do 4 times 6 side giving us a total of 24cm2.

To find the total surface area of the 8 individual cubes, we multiply 24cm2 by 8 to give us a total of 192cm2.

Now to find the total surface area of the one large cube, we know that each side of one of the small cubes is 4cm2 and the large cube is set up so that there are two levels of four cubes right on top of each other. So, the total area of each side of the large cube is 4cm2 times 4 which gives us 16cm2.

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96:192

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While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the
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Distinguish between a converging lens and a diverging lens.
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Explanation:

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3 years ago
Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur
Yanka [14]

Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

\mu_{J,T}=(\frac{dT}{dP})_H

or,

\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}

As per question the formula will be:

\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}   .........(1)

where,

\mu_{J,T} = Joule-Thomson coefficient of the gas = 0.13K/atm

T_1 = initial temperature = 19.0^oC=273+19.0=292.0K

T_2 = final temperature = ?

P_1 = initial pressure = 200.0 atm

P_2 = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}

T_2=266.12K

Therefore, the final temperature of gas is 266.12 K

5 0
3 years ago
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