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Novosadov [1.4K]

3 years ago

7

How much heat h1 is transferred to the skin by 25.0 g of steam onto the skin? the latent heat of vaporization for steam is l=2.2

56×106j/kg.express the heat transferred, in joules, to three significant figures?

1 answer:

elena-s [515]3 years ago

8 0

The heat transferred by the steam to the skin is given by
$Q=m L_v$
where
m is the mass of the steam
$L_v$ is the latent heat of vaporization.

In our problem, the mass of the steam is (converting into kg)
$m=25.0 g=0.025 kg$
while the latent heat of vaporization of the steam is
$L_v = 2.256 \cdot 10^6 J/kg$
Substituting into the previous formula, we find the heat transferred to the skin:
$Q=m L_v = (0.025 kg)(2.256 \cdot 10^6 J/kg)=56400 J = 2.56 \cdot 10^4 J$

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Read 2 more answers

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

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