Calculating Temperature and Wavelength of Sound with Distance

[MD2000]

A 93 g piece of copper is heated in a furnace to a temperature T. The copper is then inserted into a 150 g copper calorimeter containing 203 g of water. The initial temperature of the water and calorimeter is 16°C, and the final temperature after equilibrium is established is 38°C. When the calorimeter and its contents are weighed, 1.2 g of water are found to have evaporated. What was the temperature T?

For this one I basically used the following:
MCT (water) + MCT(Calo.) = - MCT(Copper)
(.150)(4186)(38-16) + (.203)(387)(38-16) = - (.093)(387)(38-Tf)

I then solved for Tf and got 469.72..which is wrong..what am I doing wrong?

------------------------------------A sound source emits sounds of frequency 185 Hz that travel through still air at 340 m/s. The source moves away from the stationary listener at 75 m/s. Find the wavelength of the sound waves between the source and the listener. X

For this I used

F prime = F (1/(1+Vs/V)
After getting F prime to equal 151.6..I plugged it into (Wavelength = V/F) the equation and got .446 m for the answer but that's also wrong..: \

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By how many decibels do you reduce the sound intensity level due to a source of sound if you double your distance from it? Assume that the waves expand spherically.

Finally on this one I used:

I = P/A for both scenarios and noticed that being farther meant a .25 difference in the sound..

any help guys?

 

[Astronuc]

1.2 g of water are found to have evaporated

In the first problem, one needs to consider that the water evaporated - or was converted to water vapor (steam), and that amount of energy has to be considered with the enthalpy of the water and calorimeter.

Here is a reference on the Doppler effect (http://hyperphysics.phy-astr.gsu.edu/hbase/sound/dopp.html) - please check your work.

 

[MD2000]

Hmm..so does the following make sense? Where the m of the steam is the m of water minus 1.2? I'm kinda confused as to why we do that tho..

MCT (water) + MCT(Calo.) + MCT (steam) = MCT(Copper)

As for the doppler one I re-did it and got 2.58 m..

What about the fourth one..what am I doing wrong?

 

[Astronuc]

Given in the problem: "The copper is then inserted into a 150 g copper calorimeter containing 203 g of water. "

MCT (water) + MCT(Calo.) = - MCT(Copper)
(.150)(4186)(38-16) + (.203)(387)(38-16) = - (.093)(387)(38-Tf)

Here one has the value of specific heat of water with the mass of copper and specific heat of copper with mass of water. So one needs to check one's work to be sure to use the correct specific heats with corresponding masses.

1.2 g of water evaporated, so one must use the heat of vaporization.
http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/phase2.html#c3

That leaves 201.8 g of water in the calorimeter.

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In the last one, what is the equation (formula) for decibels?

 

[MD2000]

Ahh..I always make stupid mistakes..so since its vaporizing..I need to use mLv on the right side of the equation to account for the vaporization of the fluid?

 

[Astronuc]

MCT (water) + MCT(Calo.) + MCT (steam) = MCT(Copper)

This is correct, but one has to be careful with the signs, or temperature differences, and appropriate masses and specific heats (and heat of fusion).

The heat 'from' the 93 g piece of copper flows into the water (some of which becomes steam) and 150 g copper calorimeter. One is basically doing a energy balance.

 

[MD2000]

Hmm..

MCT (water) + MCT(Calo.) + MCT(steam) MLV (steam) = MCT(Copper)
(.203)(4186)(38-16) + (.150)(387)(38-16) + (.2018)(1970)(38-16) + (.2018)(33.5x10^4) = (.093)(387)(38-Tf)

And simply solving for Tf?

 

[Astronuc]

Don't include the temperature of the steam in the equation, just the heat of vaporization, which is lost from the water.

 

[MD2000]

Ahh..gotcha..thanks..

what about the third one? Did I simply ratio it wrong? Is it 4 instead of .25?

 

[MD2000]

hmm..when I solve for T I'm getting a huge number..

MCT (water) + MCT(Calo.) + MLV (steam) = MCT(Copper)
(.203)(4186)(38-16) + (.150)(387)(38-16) + (.2018)(22.4x10^5) = (.093)(387)(38-Tf)

any idea what I am doing wrong?

 

[MD2000]

Hmm..so I've tried a couple different methods and am getting a huuge answer..not sure what I'm doing wrong..any help with this or the third one anyone?

 

[Hootenanny]

For the third question note that the decibel scale is defined thus;

$$I(dB) = 10\log_{10}\left(\frac{I}{I_{0}}\right)$$

Where I₀ is the threshold of hearing.

 

[Astronuc]

(.203)(4186)(38-16) + (.150)(387)(38-16) + (.2018)(22.4x10^5) = (.093)(387)(38-Tf)

The steam is 1.2 g, while 201.8 g of water remain.

and for the copper block, the temperature difference in the equation should be (Tf-38).

 

[thiotimoline]

1.2g of water evaporated actually comes from 203g water initially.