long does it take to boil away 2.40 kg of the liquid.
Boiling point of He is 
Latent heat of vapourization 
Power of electrical heater 
mass of liquid is 
amount of heat required to boil
Power 
The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.
The heat that is used or lost as matter melts and transitions from a solid to a fluid form at a constant temperature is known as the latent heat of fusion.
Due to the fact that during softening the heat energy anticipated to transform the substance from solid to fluid at air pressure is the latent heat of fusion and that the temperature remains constant during the process, the "enthalpy" of fusion is a latent heat. The enthalpy change of any quantity of material during dissolution is known as the latent heat of fusion.
For learn more about Latent heat of vaporization, visit: brainly.com/question/14980744
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To solve the problem it is necessary to apply the Malus Law. Malus's law indicates that the intensity of a linearly polarized beam of light, which passes through a perfect analyzer with a vertical optical axis is equivalent to:
Where,
indicates the intensity of the light before passing through the polarizer,
I is the resulting intensity, and
indicates the angle between the axis of the analyzer and the polarization axis of the incident light.
Since we have two objects the law would be,
Replacing the values,
Therefore the intesity of the light after it has passes through both polarizers is 
Answer:
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
Explanation:given values
Half life of lipase t_1/2 = 8 min x 60s/min = 480 s
Rate constant for first order reaction
k_d = 0.6932/480 = 1.44 x 10^-3 s-1
Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L
rate of hydrolysis Vm0 = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3
K_m = 5mmol/L
time take is given by
![t= -\frac{1}{K_d}ln[1-\frac{K_d}{V_m_0}(k_mln\frac{s_0}{s}+(s_0-s))]](https://tex.z-dn.net/?f=t%3D%20-%5Cfrac%7B1%7D%7BK_d%7Dln%5B1-%5Cfrac%7BK_d%7D%7BV_m_0%7D%28k_mln%5Cfrac%7Bs_0%7D%7Bs%7D%2B%28s_0-s%29%29%5D)
all values are given and putting these value we get
t=1642.83 secs
which is equal to
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
