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2 years ago

What mass of ice (in g) can be melted if 27.2 kJ of thermal energy are added at the freezing point? Use molar mass = 18.02 g/mol

Physics

1 answer:

san4es73 [151]2 years ago

3 0

Answer : The mass of ice melted can be, 3.98 grams.

Explanation :

First we have to calculate the moles of ice.

$Q=\frac{\Delta H}{n}$

where,

Q = energy absorbed = 27.2 kJ

$\Delta H$ = enthalpy of fusion of ice = 6.01 kJ/mol

n = moles = ?

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

$27.2kJ=\frac{6.01kJ/mol}{n}$

$n=0.221mol$

Now we have to calculate the mass of ice.

$\text{Mass of ice}=\text{Moles of ice}\times \text{Molar mass of ice}$

Molar mass of ice = 18.02 g/mol

$\text{Mass of ice}=0.221mol\times 18.02g/mol=3.98g$

Thus, the mass of ice melted can be, 3.98 grams.

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$L_A = L_B =L$ same length

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We also know that $r_A = 2 r_B$ where r represent the radius.

Since we know that a wire have a cylindrical form we can find the area for each case:

$A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B$

$A_B = \pi r^2_B$

So then we have that $A_A = 4 A_B$

Now we know that from the definition the drift velocity of electron in a wire is given by:

$v_d = \frac{I}{neA}$

Where I is the current, n the number of electrons per cubic meter, e is the charge for the electron and A the area.

If we replace we have this:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

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$V_A = \frac{1}{4} V_B$

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a. vA = vB/4

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