A few questions about water pressure, temp, and sound

AI Thread Summary
The discussion revolves around solving physics problems related to sound waves and thermal equilibrium. The first problem involves calculating the wavelength of sound emitted by a source moving away from a listener, with the correct answer being 2 meters. The second problem estimates the time required to heat water using a 380-watt immersion heater, yielding a time of approximately 87.25 seconds. The main focus is on determining the amount of ice in a mixture after a copper block is added, with calculations showing that 0.36 kg of ice was initially present. The discussion emphasizes the application of heat transfer principles in solving these problems.
Scirel
Messages
10
Reaction score
0
OK.

I have done most of them correctly on my own, I need 2 checked and help with one:

Check:

A sound source emits sounds of frequency 210 Hz that travel through still air at 340 m/s. The source moves away from the stationary listener at 80 m/s. Find the wavelength of the sound waves between the source and the listener.

I got Wavelength =2m

A small immersion heater is rated at 380 Watts. Estimate how long it will take to heat a cup of soup (assume this is 240g of water) from 23oC to 56oC

I got t=87.2451s


Now I need help with this one:

A 3.8 kg block of copper at a temperature of 79oC is dropped into a bucket containing a mixture of ice and water whose total mass is 1.2 kg. When thermal equilibrium is reached the temperature of the water is 8°C. How much ice was in the bucket before the copper block was placed in it?
(Neglect the heat capacity of the bucket.)

I know heat gained=heat lost, so:

3.8*(79-8)*387(heat capacity of copper) = ?

I don`t know how to treat the mixture of ice and water. can someone please help me?
 
Last edited:
Physics news on Phys.org
Please! I truly need help before 10:00!
 


Sure, I can help you with this problem. Let's break it down step by step.

First, let's calculate the heat lost by the copper block as it cools down from 79°C to 8°C. We can use the formula Q = mcΔT, where Q is the heat lost, m is the mass of the copper block, c is the specific heat capacity of copper, and ΔT is the change in temperature.

So, Q = (3.8 kg)(387 J/kg°C)(79°C-8°C) = 106,666.8 J

Next, we need to calculate the heat gained by the ice and water as they warm up from 0°C to 8°C. Since we know that the final temperature of the mixture is 8°C, we can use the formula Q = mcΔT again, but this time we need to use the specific heat capacity of water. The total mass of the ice and water is 1.2 kg, so:

Q = (1.2 kg)(4186 J/kg°C)(8°C-0°C) = 40,224 J

Now, since we know that heat gained = heat lost, we can set these two equations equal to each other and solve for the mass of ice in the bucket.

106,666.8 J = 40,224 J + (m)(334 J/kg°C)(8°C-0°C)

Solving for m, we get m = 0.36 kg of ice.

Therefore, there was 0.36 kg of ice in the bucket before the copper block was placed in it. I hope this helps! Let me know if you need any further clarification.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top