Line d represents water. if the atmospheric pressure in a flask is lowered to 70 kpa, water would boil at what temperature?

Physics

1 answer:

sp2606 [1]3 years ago

5 0

The diagram is in the picture attached.
Options are:
A) 32 °C
B) 70 °C
C) 92 °C
D) 100 °C

In order to find the value required, you need to look at the diagram and follow these steps:
1) search for the value of 70 kPa on the y-axis;
2) move on a horizontal line towards the right until you reach the line D;
3) move on a vertical line down, towards the x-axis;
4) read at what value of °C you are at.

Doing so, you can see that you are at a value a little bit above 90 °C (see picture).

Hence, the correct answer is C) 92°C.

You might be interested in

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli

serious [3.7K]

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Explanation:

Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

$p_{ox} = p_{oAx} + p_{oBx} (1)$

We can do exactly the same for the initial momentum along the y-axis:

$p_{oy} = p_{oAy} + p_{oBy} (2)$

The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

$p_{fx} = (m_{A} + m_{B} ) * v_{fx} (3)$

We can repeat the process for the y-axis, as follows:

$p_{fy} = (m_{A} + m_{B} ) * v_{fy} (4)$

Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

$v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} = 2.96 m/s (5)$

In the same way, we can find the component of the final momentum along the y-axis, as follows:

$v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} = 3.15 m/s (6)$

With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

$v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)$

The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

$p_{f} = (m_{A} + m_{B})* v_{f} = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)$

Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
We can find this angle applying the definition of tangent of an angle, as follows:

$tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)$