Explanation:
According to the psychrometric chart at dry bulb temperature of 
and RH 70%, the absolute humidity = 0.011 kg/kg dry air
Formula to calculate humid volume is as follows.
dry air = 
= 
= 
= 0.847 
dry air
Hence, humid volume of air is 0.847 dry air.
Specific enthalpy of dry air = specific heat capacity of dry air × dry bulb temperature
= 
= 21.126 kJ/kg dry air
Hence, the specific enthalpy of the air is 21.126 kJ/kg dry air.
As per the psychrometric chart at given conditions wet bulb temperature = 17.5 
As per the psychrometric chart at given conditions dew point temperature = 15.5
Liquid 2 because the lower the density the more it floats and the higher the density the more it sinks. The order from top to bottom is liquid 2, liquid 3, liquid 1
This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.
The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.
However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.
As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:
First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:
Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:
This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.
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Answer:
i dont know what you are saying
Explanation:
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