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fiasKO [112]
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.
Physics
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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A monk is sitting atop a mountain in complete rest in meditation. What is the kinetic Energy of the monk? (assume mass of 65 kg
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Two long, parallel wires carry currents of different magnitudes. If the amount of current in one of the wires is doubled, what h
Ulleksa [173]

Answer:

The magnitude of the force that each wire exerts on the other will increase by a factor of two.

Explanation:

force on parallel current carrying wire, F = BILsinθ

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I is the magnitude of current on the wire

θ is the angle of inclination of the wire

Assuming B, L and θ is constant, then F ∝ I

F = kI

\frac{F_1}{I_1} = \frac{F_2}{I_2}

When the amount of current is doubled in one of the wires, lets say the second wire;

\frac{F_1}{I_1} = \frac{F_2}{2I_1} \\\\F_2 = \frac{2F_1I_1}{I_1} \\\\F_2 =2F_1

Also, if will double the amount of current on the first wire, then

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Answer:

Explanation:

From the information given,

V = 4 volts

A = 2 amps

a)

In the first instance, one light on the sting goes out and the whole string of lights no longer turns on. This means that the circuit is a series circuit.

b) Total voltage = 4 x 100 = 400 V

The current passing through each bulb is the same. Thus

Total Current = 2 Amps

Recall, V = IR

R = V/I

Thus,

Resistance = 400/2

Resistance = 200 ohms

c) In this case, one light goes out on the string all other lights still turn on. This means that the circuit is a parallel circuit

d)

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Total resistance = 0.02 ohms

3 0
1 year ago
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