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professor190 [17]

3 years ago

7

Two 84.5 g ice cubes are dropped into 30 g of water in a glass. If the water is initially at a temperature of 50 C and if the ic

e comes directly from the freezer at -300 C, what will be temperature of the drink when the ice and the water reach thermal equilibrium ? How much ice and how much water are present at thermal equilibrium ? Ignore the heat capacity of the glass and heat transferred to and from the environment.

1 answer:

Setler [38]3 years ago

4 0

Answer:

final temperature will be 0 degree C

Total amount of ice will be

$m_{ice} = 182 g$

total amount of water

$m_{water} = 17 g$

Explanation:

After thermal equilibrium is achieved we can say that

Heat given by water = heat absorbed by ice cubes

so we will have

Heat given by water to reach 0 degree C

$Q_1 = m_1s_1 \Delta T_1$

$Q_1 = 0.030(4186)(50 - 0)$

$Q_1 = 6279 J$

heat absorbed by ice cubes to reach 0 degree

$Q_2 = m_2 s_2 \Delta T_2$

$Q_2 = (0.169)(2100)(30)$

$Q_2 = 10647 J$

so we will have

$Q_2 > Q_1$

so here we can say that few amount of water will freeze here to balance the heat

$10647 - 6279 = mL$

$m = \frac{10647 - 6279}{335000}$

$m = 13 g$

so final temperature will be 0 degree C

Total amount of ice will be

$m_{ice} = 84.5 + 84.5 + 13$

$m_{ice} = 182 g$

total amount of water

$m_{water} = 30 - 13$

$m_{water} = 17 g$

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Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

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Explanation:

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Well, according to the law of universal gravitation:

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Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

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$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

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3 years ago

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Answer:

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Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

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Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

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