7) Specific latent heat of evaporation: [tex]8.5\cdot 10^5 J/kg[/tex]
8a) Time needed: 28 s
8b) Time needed: 181 s
Explanation:
7)
When a substance at boiling temperature is supplied with heat, the amount of substance that completely boils is given by the equation
[tex]Q=\Delta m L_v[/tex]
where
Q is the amount of heat supplied
[tex]\Delta m[/tex] is the amount of substance that boils
[tex]L_v[/tex] is the specific latent heat of evaporation
In this problem, we have:
[tex]Q=15.3 kJ = 15300 J[/tex] is the amount of heat supplied
[tex]\Delta m=0.25 kg - 0.232 kg =0.018 kg[/tex] is the amount of substance that has boiled
Therefore, the specific latent heat of evaporation of the substance is:
[tex]L_v=\frac{Q}{\Delta m}=\frac{15300}{0.018}=8.5\cdot 10^5 J/kg[/tex]
8a)
The amount of heat energy needed to increase the temperature of a certain substance is given by
[tex]Q=mC\Delta T[/tex]
where
m is the mass of the substance
C is its specific heat capacity
[tex]\Delta T[/tex] is the change in temperature
In this problem, we have:
m = 200 g = 0.2 kg is the mass of water
[tex]C=4186 J/kg^{\circ}C[/tex] is the specific heat capacity of water
[tex]\Delta T = 100-16=84^{\circ}C[/tex] is the change in temperature
Substituting,
[tex]Q=(0.2)(4186)(84)=70,325 J[/tex]
The power of the kettle is
P = 2500 W
Therefore, the time needed is:
[tex]t=\frac{Q}{P}=\frac{70325}{2500}=28 s[/tex]
8b)
The amount of heat energy needed in order to boil the water away is
[tex]Q=mL_v[/tex]
where
m is the mass of the water
[tex]L_v[/tex] is the specific latent heat of evaporation of water
Here we have:
m = 200 g = 0.200 kg
[tex]L_v=2.26\cdot 10^6 J/kg[/tex]
Therefore,
[tex]Q=(0.200)(2.26\cdot 10^6)=4.52\cdot 10^5 J[/tex]
The power of the kettle is
P = 2500 W
Therefore, the time needed to boil the water is:
[tex]t=\frac{Q}{P}=\frac{4.52\cdot 10^5}{2500}=181 s[/tex]
Learn more about specific heat capacity:
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