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Biology; Chemistry; Physics: Post your doubts here!

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now we are going off topic..
can anyone tell me how to do this...:(

Water at a temperature of 16 °C enters an ice-making machine and emerges as ice cubes at a
temperature of –5 °C. The melting point of ice is 0 °C.
(a) Calculate the total energy removed from 1.0 kg of water as it cools from 16 °C, changes into
ice, and then cools to –5 °C.
specific heat capacity of liquid water = 4.2 × 103 J / (kg °C)
specific latent heat of fusion of water = 3.4 × 105 J / kg
specific heat capacity of ice = 2.1 × 103 J / (kg °C)
energy

The answer which I got id 8820 J
tell me if I'm correctso i will tell you the method
 
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just suggestion it is due to in solid state the bond between molecules are not much stronger than water in liquid!!!
That seems to be inaccurate.

suggest why less energy is needed to change ice into water than to change the same
mass of water into steam
this too
Because the amount of energy required by molecules to break away completely from inter-molecular forces and escape into the air in gaseous form is more (as a fact). Whereas, energy required to just dissemble the arrangement of atoms in a solid slightly is less :D my guess. :p
 
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That seems to be inaccurate.


Because the amount of energy required by molecules to break away completely from inter-molecular forces and escape into the air in gaseous form is more (as a fact). Whereas, energy required to just dissemble the arrangement of atoms in a solid slightly is less :D my guess. :p

seems much better...you sure?? :D
 
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.
can anyone tell me how to do this...:(

Water at a temperature of 16 °C enters an ice-making machine and emerges as ice cubes at a
temperature of –5 °C. The melting point of ice is 0 °C.
(a) Calculate the total energy removed from 1.0 kg of water as it cools from 16 °C, changes into
ice, and then cools to –5 °C.
specific heat capacity of liquid water = 4.2 × 103 J / (kg °C)
specific latent heat of fusion of water = 3.4 × 105 J / kg
specific heat capacity of ice = 2.1 × 103 J / (kg °C)
energy

Which Year's paper is it?
 
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Okay I got it
The answer has to be solved in two steps
-First calculate the energy lost from converting water to ice at 0 degrees=> E=mc(T1-T2)=> 4.2x103 x 1.0kg x (16-0)=> 67200J
-Next calculate the energy lost from conversion of ice at 0 degrees to at -5 degrees=> mc(T1-T2)=>1.0kg x 2.1x103 x [0-(-5)]=> 10500J
Add them together=> 67200J+10500J=>77700J
 
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Okay I got it
The answer has to be solved in two steps
-First calculate the energy lost from converting water to ice at 0 degrees=> E=mc(T1-T2)=> 4.2x103 x 0.1kg x (16-0)=> 6720J
-Next calculate the energy lost from conversion of ice at 0 degrees to at -5 degrees=> mc(T1-T2)=>0.1kg x 2.1x103 x [0-(-5)]=> 1050J
Add them together=> 6720J+1050J=>7770J

check the marking scheme the answer is something like 4.2x10^5
i think please check this is incorrect!! :D
 
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