- #1
mopar969
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Here is the help I was given on a later post but the thread is dead so I am reposting here:
Lets first think about the ice. The energy gained from the aluminium cup and the water will be used for heating ice. Let's first find out how much energy is required to completely melt the ice.
Specific heat of ice = 2090 J/kg-C = cice
Latent heat of fusion for ice = LaTeX Code: 3.33 \\times 10^5 J/kg =Lf
mass of ice = 0.02 kg = mice
so amount of energy required to completely melt the ice is
LaTeX Code: Q_{ice}= m_{ice}\\left[ c_{ice}(\\Delta T) +L_f \\right]
where LaTeX Code: \\Delta T = 10^{\\circ} C = 0^{\\circ}C - (-10^{\\circ} C)
I got Qice = 7078 J.
Now specific heat of aluminium = cAl = 900 J/kg-C
specific heat of water = cw = 4186 J/kg-C
mass of Aluminium = mAl = 0.1 kg
mass of water = mw = 0.2 kg
Now if the aluminium cup + water were to come to the 0 celcius, the amount of energy which will be released is
LaTeX Code: Q_2 = m_{Al}c_{Al}(25-0) + m_w c_w (25-0)
which is eqaul to 23180 J. But we need only 7078 J to completely melt the ice. So
rest of the energy (23180- 7078 = 16102 J) will be used to increase the temperature
of the melted water. Let's say the equivalent temperature of the whole thing is T. so we will now write the equation where left hand side is the total energy gained by the ice and the right hand side is the total energy lost by Aluminium + water
LaTeX Code: 7078 + m_{ice} c_w (T-0) = m_{Al} c_{Al} (25 -T) + m_w c_w (25-T)
solving which I got LaTeX Code: T = 15.92^{\\circ} C
Here is a link to the post
https://www.physicsforums.com/showthread.php?t=467341
Anyhow I don't understand how the answer for the last equation was solved please help me solve the equation
Lets first think about the ice. The energy gained from the aluminium cup and the water will be used for heating ice. Let's first find out how much energy is required to completely melt the ice.
Specific heat of ice = 2090 J/kg-C = cice
Latent heat of fusion for ice = LaTeX Code: 3.33 \\times 10^5 J/kg =Lf
mass of ice = 0.02 kg = mice
so amount of energy required to completely melt the ice is
LaTeX Code: Q_{ice}= m_{ice}\\left[ c_{ice}(\\Delta T) +L_f \\right]
where LaTeX Code: \\Delta T = 10^{\\circ} C = 0^{\\circ}C - (-10^{\\circ} C)
I got Qice = 7078 J.
Now specific heat of aluminium = cAl = 900 J/kg-C
specific heat of water = cw = 4186 J/kg-C
mass of Aluminium = mAl = 0.1 kg
mass of water = mw = 0.2 kg
Now if the aluminium cup + water were to come to the 0 celcius, the amount of energy which will be released is
LaTeX Code: Q_2 = m_{Al}c_{Al}(25-0) + m_w c_w (25-0)
which is eqaul to 23180 J. But we need only 7078 J to completely melt the ice. So
rest of the energy (23180- 7078 = 16102 J) will be used to increase the temperature
of the melted water. Let's say the equivalent temperature of the whole thing is T. so we will now write the equation where left hand side is the total energy gained by the ice and the right hand side is the total energy lost by Aluminium + water
LaTeX Code: 7078 + m_{ice} c_w (T-0) = m_{Al} c_{Al} (25 -T) + m_w c_w (25-T)
solving which I got LaTeX Code: T = 15.92^{\\circ} C
Here is a link to the post
https://www.physicsforums.com/showthread.php?t=467341
Anyhow I don't understand how the answer for the last equation was solved please help me solve the equation