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3 years ago

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50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam

container so that heat loss can be ignored. Find the final temperature.

1 answer:

Arada [10]3 years ago

5 0

Answer : The final temperature is, $25.0^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 50 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $-15^oC$

$T_2$ = initial temperature of water = $30^oC$

Now put all the given values in the above formula, we get:

$50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC$

$T_f=25.0^oC$

Therefore, the final temperature is, $25.0^oC$

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Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

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Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

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but,

K.E = (1/2) mv²

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Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

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3 years ago

How often must you check the temperature of food that is being held with temperature control?.

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1 year ago

In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for

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Answer:

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Explanation:

From the question we are told that

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The potential across the capacitor can be mathematically represented as

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Where $V_o$ is the voltage of the capacitor when it is fully charged

So at $\tau = 3$

$V = V_o (1 - e^{- 3})$

$V = 0.950213 V_o$

Generally energy stored in a capacitor is mathematically represented as

$E = \frac{1}{2 } * C * V ^2$

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now since capacitance is constant at $\tau = 3$

The energy stored can be evaluated at as

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Answer:

The minimum wall thickness required for the spherical tank is 0.0189 m

Explanation:

Given data:

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