How much heat is needed to boil 120 kg of water ?

1 answer:

7 0

Q = mc∆t, where:
q = energy flow
m = mass, 120 000 g
c = specific heat capacity, 4.81 J/gC
∆t = change in temperature, ~75 (100 - 25, which is room temperature)

Substituting in the values, we get:
q = 120000 x 4.81 x 75 = 43290000 Joules = 43.29 MJ

Hope I helped!! xx

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Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

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Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$