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laila [671]
3 years ago
9

How much energy does it take to melt a 16.87 g ice cube? ΔHfus = 6.02 kJ/mol How much energy does it take to melt a 16.87 g ice

cube? = 6.02 kJ/mol 108 kJ 102 kJ 5.64 kJ 936 J none of the above
Physics
1 answer:
Vilka [71]3 years ago
5 0

Answer:

How much energy does it take to melt a 16.87 g ice cube? ΔHfus = 6.02 kJ/mol How much energy does it take to melt a 16.87 g ice cube? = 6.02 kJ/mol

A. 108 kJ

B. 102 kJ

C. 5.64 kJ

D. 936 kJ

E. none of the above

<em>5.64 kJ</em>

Explanation:

The Heat of fusion is the heat energy required to dissolve a given mass of ice at melting point.

<h3>Step by Step Calculation</h3>

The heat energy required to dissolve ice can be calculated using the expression below;

Q = ΔH_{f} x m ...............................................1

where Q is the heat energy required;

           ΔH_{f}  is the heat of fusion for ice;

           m is the mole

All the parameters above are provided in the question except m, so to get m we use the molar mass of water (also for ice) which is 18.01528 g/mol .

<em>This means that 18.01528 g of ice is contained in one mole, therefore the mole for 16.87 g of ice is given as;</em>

m = \frac{16.87g}{18.015g/mol}

m = 0.9364 mole of ices

Now the parameters are complete, we are given;

ΔH_{f}  = 6.02 kJ/mol

m = 0.9364 mol

Q =?

Substituting into equation 1, we have

Q =  6.02 kJ/mol x 0.9364 mol

Q = 5.64 kJ

<em>Therefore, the energy required to melt 16.87 g of ice is 5.64 kJ</em>

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Current Flow and Ohm's Law

Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current, voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.

I = V / R 

Where: 

I =

Electrical Current (Amperes)

V =

Voltage (Voltage)

R =

Resistance (Ohms)

    

Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.

The Java applet below allows the user to vary each of these three parameters in Ohm's Law and see the effect on the other two parameters. Values may be input into the dialog boxes, or the resistance and voltage may also be varied by moving the arrows in the applet. Current and voltage are shown as they would be displayed on an oscilloscope with the X-axis being time and the Y-axis being the amplitude of the current or voltage. Ohm's Law is valid for both direct current (DC) and alternating current (AC). Note that in AC circuits consisting of purely resistive elements, the current and voltage are always in phase with each other.

Exercise: Use the interactive applet below to investigate the relationship of the variables in Ohm's law. Vary the voltage in the circuit by clicking and dragging the head of the arrow, which is marked with the V. The resistance in the circuit can be increased by dragging the arrow head under the variable resister, which is marked R. Please note that the vertical scale of the oscilloscope screen automatically adjusts to reflect the value of the current.

See what happens to the voltage and current as the resistance in the circuit is increased. What happens if there is not enough resistance in a circuit? If the resistance is increased, what must happen in order to maintain the same level of current flow?


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F should be 10. If The Whole top is 50cm, then we should subtract 10n and 30n which gives us 10.

Or it could be 15 if both top and bottom are 25. 10+15= 25.
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