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ivolga24 [154]
3 years ago
5

How long does it take a 750-W coffeepot to bring to a boil 0.65 L of water initially at 13 ∘C? Assume that the part of the pot w

hich is heated with the water is made of 360 g of aluminum, and that no water boils away. Ignore the heat loss to the surrounding environment. The value of specific heat for water is 4186 J/kg⋅C∘ and for aluminum is 900 J/kg⋅C∘.
Physics
1 answer:
aksik [14]3 years ago
6 0

Answer:

353 s

Explanation:

Energy               = energy needed to raise the temperature of pot to 100°C +

required              energy needed to raise the temperature of water to 100°C

                    = MCΔT + mcΔT                      

                     = 0.65 kg * 4186 J/kg-°C * 87 °C + 0.36 kg * 900 J/kg-°C * 87 °C

                     = 264 906.3 J

Energy required = energy delivered * time

264 906.3 J = 750 J/s * time

time = 353 s

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<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles.

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Dillon wants to go to the park to play basketball. To get to the park, he must travel 9 km East, and then 6 km South, then 9 Km
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Answer:

24 km

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From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s
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Solution:

Let the slope of the best fit line be represented by 'm_{best}'

and the slope of the worst fit line be represented by 'm_{worst}'

Given that:

m_{best} = 1.35 m/s

m_{worst} = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

\Delta m = \frac{m_{best}-m_{worst}}{2}               (1)

Substituting values in eqn (1), we get

\Delta m = \frac{1.35 - 1.29}{2} = 0.03 m/s

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Distance measurements based on the speed of light; used for objects in space
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In a one year light travels 9460000000000 kilometer.

We know that, speed of light is 3\times10^{8}\ m/s

and  time is 31536000 seconds in 1 year

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A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the
vladimir2022 [97]

Answer:

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table.  R = 0.287 kJ/kg-k

The Pressure-Volume relation is <em>PV</em> = <em>C</em>

<em>T = C </em> for isothermal process

Calculating for the work done in isothermal process

<em>W</em> = <em>P</em>₁<em>V</em>₁ ln[\frac{P_{1} }{P_{2} }]

   = <em>mRT</em>₁ln[\frac{P_{1} }{P_{2} }]      [∵<em>pV</em> = <em>mRT</em>]

   = (5) (0.287) (272.039) ln[\frac{2.0}{1.0}]

   = 270.588 kJ

Since the process is isothermal, Internal energy change is zero

Δ<em>U</em> = mc_{v}(T_{2}  - T_{1} ) = 0

From 1st law of thermodynamics

Q = Δ<em>U  </em>+ <em>W</em>

   = 0 + 270.588

   = 270.588 kJ

4 0
3 years ago
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