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3 years ago

Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water

Physics

1 answer:

photoshop1234 [79]3 years ago

4 0

Answer:

7.55 g

Explanation:

Given that:

Heat of fusion = 333kj/kg

Heat capacity, c = 4190 j/kg /k

The Number of grams of ice that will melt can be represented as y:

Number of grams of ice that will melt * heat of fusion = specific heat capacity * temperature change

y * 333 * 10^3 J = (4190) * (6 - 0)

333000y = 25140

y = 25140 / 333000

y = 0.0754954 kg

y = 0.0754954 * 100

y = 7.549 g

Hence, Number of grams of ice that will melt = 7.55 g

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Explanation:

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3 years ago

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag

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Answer:

(a)0.0675 J

(b)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

$E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675 J$

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

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3 years ago

Newton’s Laws of Motion are absolute in classical physics. One example that uses all three laws simultaneously is the firing of

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3 years ago

Read 2 more answers

(a) What is the entropy change of a 14.6 g ice cube that melts completely in a bucket of water whose temperature is just above t

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Answer:

a) 17.81 J/K

b) 33.325 J/K

Explanation:

The expression to use here is the following:

ΔS = Q/T

Where:

Q: heat released or absorbed

T: Temperature in K

Now, in order to do this, we need to gather the data. We know that the temperature in part a) is above the freezing temperature of water, which is 0 ° C or 273 K. and the mass of the ice cube is 14.6 g.

a) Using the water heat of fusion (Cause it's melting), we can calculated the heat released using the following expression:

Q = m * Lf

Lf = 333,000 J/kg

Solving for Q first we have:

Q = (14.6 / 1000) * 333,000

Q = 4,861.8 J

Now, the entropy change is:

ΔS = 4,861.8 / 273

ΔS = 17.81 J/K

b) In this part, we follow the same procedure than in part a) but using the water heat of boiling (Lv = 2,256,000 J/kg), the temperature of boiling which is 100 °C (or 373 K) and the mass of 5.51 g (0.00551 kg)

Calculating the heat:

Q = 0.00551 * 2,256,000 = 12,430.56 J

Now the entropy change:

ΔS = 12,430.56 / 373

ΔS = 33.325 J/K

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3 years ago

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4 years ago