During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t
2 answers:
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Answer : The heat change of the cold water in Joules is,
Explanation :
First we have to calculate the mass of cold water.
As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.
Now we have to calculate the heat change of cold water.
Formula used :
where,
Q = heat change of cold water = ?
m = mass of cold water = 45 g
c = specific heat of water =
= initial temperature of cold water =
= final temperature =
Now put all the given value in the above formula, we get:
Therefore, the heat change of cold water is
Answer:
Work out = 28.27 kJ/kg
Explanation:
For R-134a, from the saturated tables at 800 kPa, we get
= 171.82 kJ/kg
Therefore, at saturation pressure 140 kPa, saturation temperature is
= -18.77°C = 254.23 K
At saturation pressure 800 kPa, the saturation temperature is
= 31.31°C = 304.31 K
Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.
Thus, =
= 171.82 kJ/kg
We know COP of heat pump
COP =
=
= 6.076
Therefore, Work out put, W =
= 171.82 / 6.076
= 28.27 kJ/kg