When a solid object is completely immersed in water it has an apparent weight of 17.3 N.
When the object it is completely immersed in alcohol, which has a relative density of
0.785, its apparent weight is 22.5 N. Determine the volume of the object. [
and
A certain quantity of ice at −20
o
C is converted to steam at 100
o
C. The conversion requires
8.5 kW of power applied for a time of 1500 s. Calculate the mass of the ice.
When the object it is completely immersed in alcohol, which has a relative density of
0.785, its apparent weight is 22.5 N. Determine the volume of the object. [
and
A certain quantity of ice at −20
o
C is converted to steam at 100
o
C. The conversion requires
8.5 kW of power applied for a time of 1500 s. Calculate the mass of the ice.
-
Hi, Daisy. I see that you've answered questions on heat before. So I'd have figured you'd do the last one. But I'll cover it.
The first problem is just algebra. Given g is the acceleration of gravity, m is the object's mass, v is the object's volume, ρw is the density of water, and ρa is the density of alcohol, you have:
1. 17.3 N = m•g - v•ρw•g
2. 22.5 N = m•g - v•ρa•g
Subtracting (1) from (2) gives:
3. 22.5 N - 17.3 N = v•ρw•g - v•ρa•g = v•g•(ρw - ρa)
Solving for the volume, v, gives:
4. v = (22.5 N - 17.3 N) / [ g•(ρw - ρa) ]
You don't specify g, but it is usually taken to be about 9.8 m/s². ρa is 0.785•ρw from the problem. ρw is 1 gram/cc. So plugging all that in gives 2466 cc. You can test that by entering this into google:
5. (22.5 N - 17.3 N)/((earth's gravity)*(1 gram/cc - 0.785 gram/cc))
On the heat issue...
The total energy supplied is:
6. Q_supplied = 8.5 kW • 1500 s = 12.75 million joules
Some details are needed about water:
specific heat capacity of ice: c_ice = 2.108 J/g•K
specific heat capacity of water: c_water = 4.187 J/g•K
latent heat of melting: L_melt = 334 J/g
The energy equation now is:
7. Q_supplied = Q_boil + Q_ice + Q_latent
Where,
8. Q_latent = m • L_melt
9. Q_ice = m • c_ice • ( 0C - (-20C) )
10. Q_boil = m • c_water • ( 100C - (0C) )
So, grabbing up (7) above, substituting in (8), (9), and (10) above, and solving for m, gives:
11. m = Q_supplied / [ c_water • ( 100C - (0C) ) + c_ice • ( 0C - (-20C) ) + L_melt ]
Or,
12. m = 12.75•10^6 J / [ 4.187 J/g•K • 100K + 2.108 J/g•K • 20 K + 334 J/g ]
That is 16.041 kg or 16.041 liters of ice. But rounded, the answer is probably just 16 liters.
The first problem is just algebra. Given g is the acceleration of gravity, m is the object's mass, v is the object's volume, ρw is the density of water, and ρa is the density of alcohol, you have:
1. 17.3 N = m•g - v•ρw•g
2. 22.5 N = m•g - v•ρa•g
Subtracting (1) from (2) gives:
3. 22.5 N - 17.3 N = v•ρw•g - v•ρa•g = v•g•(ρw - ρa)
Solving for the volume, v, gives:
4. v = (22.5 N - 17.3 N) / [ g•(ρw - ρa) ]
You don't specify g, but it is usually taken to be about 9.8 m/s². ρa is 0.785•ρw from the problem. ρw is 1 gram/cc. So plugging all that in gives 2466 cc. You can test that by entering this into google:
5. (22.5 N - 17.3 N)/((earth's gravity)*(1 gram/cc - 0.785 gram/cc))
On the heat issue...
The total energy supplied is:
6. Q_supplied = 8.5 kW • 1500 s = 12.75 million joules
Some details are needed about water:
specific heat capacity of ice: c_ice = 2.108 J/g•K
specific heat capacity of water: c_water = 4.187 J/g•K
latent heat of melting: L_melt = 334 J/g
The energy equation now is:
7. Q_supplied = Q_boil + Q_ice + Q_latent
Where,
8. Q_latent = m • L_melt
9. Q_ice = m • c_ice • ( 0C - (-20C) )
10. Q_boil = m • c_water • ( 100C - (0C) )
So, grabbing up (7) above, substituting in (8), (9), and (10) above, and solving for m, gives:
11. m = Q_supplied / [ c_water • ( 100C - (0C) ) + c_ice • ( 0C - (-20C) ) + L_melt ]
Or,
12. m = 12.75•10^6 J / [ 4.187 J/g•K • 100K + 2.108 J/g•K • 20 K + 334 J/g ]
That is 16.041 kg or 16.041 liters of ice. But rounded, the answer is probably just 16 liters.