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3 years ago

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A 325 g iron bucket contains 527 g of water, both at 21.6°C. A 855 g piece of hot steel (T = 352 degrees Celsius) is added to bu

cket at the same time 295 g of ice at -22.2°C is added. Determine if all the ice melts. If so, what is the final temperature of the mixture? If not, how much of the ice melted?

1 answer:

Neporo4naja [7]3 years ago

3 0

Answer:

$T = 18.63 ^oC$

Explanation:

As we know that heat given by the hot piece of iron = heat absorbed by water + ice

As we know that

$c_{iron} = 0.450 J/g ^oC$

also we have

$c_{ice} = 2050 J/kg ^oC$

now we have

$325(0.450)(T-21.6) + 0.527(4186)(T - 21.6) + 0.295(2050)(0 + 22.2) + (0.295)(335000) + 0.295(4186)(T - 0) = 855(0.450)(352 - T)$

$146.25T - 3159 + 2206T - 47650 + 13425.45 + 98825 + 1234.87T = 135432 - 384.75 T$

$3971.87 T = 73990.55$

$T = 18.63 ^oC$

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The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.

<u>Solving for the amount remaining</u>

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

$y=14.2* (0.9172535661)^{(31.8)}$ $y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}$

$y=14.2* 0.0641450581$ $y=14.2*(0.5)^{3.962518068}$

$y=0.9108598257$ $y=14.2* 0.0641450581$

$y=0.9108598257$

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

8 0

2 years ago

Need today please

podryga [215]

Answer:

yeah righto the correct answer mate would be D) :)

5 0

3 years ago

Answer the amplitude part of the questions in 7, 8, & 9

erastova [34]

I actually believe for the first question, it would be complete destructive interference as the amplitude and the approximate wavelength for each are the same and will completely or entirely cancel out, rather than simply decreasing or lowering the amplitude as in the bottom question.

The amplitude for the first will be 0, as the 2 waves will cancel each other out. The amplitude of the second, will be 3x, I believe, assuming the amplitude of the first is 2x and the second is 1x, in a constructive interference, I believe the amplitudes would add up.

Likewise for the bottom, I believe you would be subtracting the supposed amplitude of the first which is 2x from 1x which would be 1x.

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4 years ago

a sealed cylinder contains a sample of ideal gas at a pressure of 2.0 atm. The rms speed of the molecules is v0. If the rms spee

fredd [130]

Answer:

P_2 = 1.62 atm

Explanation:

We know the formula for the rms speed of the ideal gas is given by

$v_{rsm}=\sqrt{\frac{3PV}{m} }$

P= pressure of the surrounding

V= volume of the vessel

m= mass of the gas

Now, From this formula rms speed (v_rms) is directly proportional to square root is pressure.

Then

$\frac{v_{rsm,1}}{v_{rsm,2}} =\sqrt{\frac{P_1}{P_2} }$

given that v_rsm,1= v0

and v_rsm,2=0.9v0

putting these values we get

$\frac{v0}{0.9v0} =\sqrt{\frac{2}{P_2} }$

P_2 = 1.62 atm

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4 years ago

Read 2 more answers

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electr

Oksi-84 [34.3K]

Answer:

1 = 5.4 J

2 = 0.1979 C

3 = 5

Explanation:

Energy in a capacitor, E is

E = 1/2 * C * V²

E = 1/2 * 3000*10^-6 * 60²

E = 1/2 * 3000*10^-6 * 3600

E = 1/2 * 10.8

E = 5.4 J

E = Q²/2C = 6.53 J

E * 2C = Q²

Q² = 6.53 * 2 * 3000*10^-6

Q² = 13.06 * 3000*10^-6

Q² = 0.03918

Q = √0.03918

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The Capacitor, C is inversely proportional to the distance of separation, D. Thus, if D is increased by 5 to be 5D, then C would be C/5. And therefore, our energy stored in the capacitor is increased by a factor of 5.

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4 years ago