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Viktor [21]
3 years ago
11

A 6.00 kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -10.0 ∘C. T

he ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 0.50 kg of ice and 1.50 kg of liquid water.
Required:
What was the initial temperature of the piece of copper?
Physics
1 answer:
julsineya [31]3 years ago
3 0

Answer:

T = 235.134^{\textdegree}C

Explanation:

The process is modelled after the First Law of Thermodynamics:

(6000\,g) \cdot \left(0.385\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (T -0^{\textdegree}C ) = (2000\,g)\cdot \left(2.108\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot [0^{\textdegree}C -(-10^{\textdegree}C)] + (1500\,g)\cdot \left(334\,\frac{J}{g} \right)

The initial temperature of the piece of solid copper is:

T = 235.134^{\textdegree}C

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Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m. If the distance between the
Aleksandr-060686 [28]

Given:

The magnitude of each charge is q1 = q2 = 1 C

The distance between them is r = 1 m

To find the force when distance is doubled.

Explanation:

The new distance is

\begin{gathered} r^{\prime}=\text{ 2r} \\ =2\times1 \\ =2\text{ }m \end{gathered}

The force can be calculated by the formula

F=k\frac{q1q2}{(r^{\prime})^2}

Here, k is the constant whose value is

k=9\times10^9\text{ N m}^2\text{ /C}^2

On substituting the values, the force will be

\begin{gathered} F=9\times10^9\times\frac{1\times1}{(2)^2} \\ =2.25\times10^9\text{ N} \end{gathered}

7 0
1 year ago
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Whitepunk [10]

Answer:

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4 0
3 years ago
The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde
Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

S_1 S_2=3m

S_1 O=4m

By Pythagoras theorem

                                 S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m

Now

The intensity at O when both speakers are on is given by

I=4I_1 cos^2(\pi \frac{\delta}{\lambda})

Here

  • I is the intensity at O when both speakers are on which is given as 6 W/m^2
  • I1 is the intensity of one speaker on which is 6  W/m^2
  • δ is the Path difference which is given as

                                           \delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m

  • λ is wavelength which is given as

                                             \lambda=\frac{v}{f}

      Here

              v is the speed of sound which is 320 m/s.

              f is the frequency of the sound which is to be calculated.

                                  16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )

where k=0,1,2

for minimum frequency f_1, k=1

                                  {f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz

So the minimum frequency is 702.22 Hz

5 0
3 years ago
Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

Mm=44.0 g/mol is the molar mass of the carbon dioxide

So, the number of moles of the gas is

n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol

Now we can find the pressure of the gas by using the ideal gas equation:

pV=nRT

where

p is the pressure

V=1.00 L = 0.001 m^3 is the volume

n = 0.023 mol is the number of moles

R=8.314 J/mol K is the gas constant

T=25.0^{\circ}+273=298 K is the temperature of the gas

Solving the equation for p, we find

p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa

And since we have

1 atm = 1.01\cdot 10^5 Pa

the pressure in atmospheres is

p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm

5 0
3 years ago
If a runner has a speed of 8.66m/s and runs for 46.2s what distance is covered? tv = d
kati45 [8]

Answer:

\text{Using the formula: }v=\frac{d}{t}\\\therefore vt=d\\\text{Plug and chug:}46.2(8.66)=400.092\text{ metres}

6 0
2 years ago
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