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3 years ago

11

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:

Speed of air = 1106.38 ft/s

Explanation:

Speed of sound in air with temperature

$v_{air}=331.3\sqrt{1+\frac{T}{273.15}} \\$

Here speed is in m/s and T is in celcius scale.

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$T=(50-32)\times \frac{5}{9}=10^0C \\$

Substituting

$v_{air}=331.3\sqrt{1+\frac{10}{273.15}}=337.31m/s \\$

Now we need to convert m/s in to ft/s.

1 m = 3.28 ft

Substituting

$v_{air}=337.31\times 3.28=1106.38ft/s \\$

Speed of air = 1106.38 ft/s

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The number of outer shell electrons determines the chemical properties of an element?

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Answer:

To calculate the tension on a rope holding 1 object, multiply the mass and gravitational acceleration of the object. If the object is experiencing any other acceleration, multiply that acceleration by the mass and add it to your first total.

Explanation:

The tension in a given strand of string or rope is a result of the forces pulling on the rope from either end. As a reminder, force = mass × acceleration. Assuming the rope is stretched tightly, any change in acceleration or mass in objects the rope is supporting will cause a change in tension in the rope. Don't forget the constant acceleration due to gravity - even if a system is at rest, its components are subject to this force. We can think of a tension in a given rope as T = (m × g) + (m × a), where "g" is the acceleration due to gravity of any objects the rope is supporting and "a" is any other acceleration on any objects the rope is supporting.[2]

For the purposes of most physics problems, we assume ideal strings - in other words, that our rope, cable, etc. is thin, massless, and can't be stretched or broken.

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3 years ago

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Answer:

The magnitude of the force is 34.59 N.

Explanation:

Given that,

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Using formula of force

$F_{net}=F_{in}-F_{out}$

$F_{net}=(P_{in}-P_{out})A$

Where, $P_{in}$ =inside Pressure

$P_{out}$ =outside Pressure

A = area

Put the value into the formula

$F_{net}=(1.90\times10^{5}-1.013\times10^{5})3.90\times10^{-4}$

$F_{net}=34.59\ N$

Hence, The magnitude of the force is 34.59 N.

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3 years ago