Change in Temperature Problem

In summary, the last equation in the conversation can be solved by using algebra to rearrange the equation and isolate the variable T, which yields a temperature of 15.92°C.
  • #1
mopar969
201
0
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
 
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  • #2
?To solve the last equation, you can use algebra to rearrange the equation and isolate the variable T on one side. First, subtract 7078 from both sides: Q2 - 7078 = mAl * cAl * (25 - T) + mw * cw * (25 - T)Then, subtract mAl * cAl * 25 from both sides: -7078 - mAl * cAl * 25 = mAl * cAl * (-T) + mw * cw * (25 - T)Next, subtract mw * cw * 25 from both sides: -7078 - mAl * cAl * 25 - mw * cw * 25 = mAl * cAl * (-T) + mw * cw * (-T)Finally, divide both sides by mAl * cAl + mw * cw: -7078 / (mAl * cAl + mw * cw) - 25 = -T / (mAl * cAl + mw * cw)You can then solve for T by multiplying both sides by (mAl * cAl + mw * cw): T = (mAl * cAl + mw * cw) * [-7078 / (mAl * cAl + mw * cw) - 25]= (0.1 * 900 + 0.2 * 4186) * [-7078 / (0.1 * 900 + 0.2 * 4186) - 25]= 15.92°C
 

1. What causes changes in temperature?

There are several factors that can cause changes in temperature, including the Earth's distance from the sun, greenhouse gas emissions, and natural climate cycles.

2. How does human activity contribute to changes in temperature?

Human activities, such as burning fossil fuels and deforestation, release large amounts of greenhouse gases into the atmosphere, trapping heat and causing the Earth's temperature to rise.

3. What are the potential consequences of changes in temperature?

Some potential consequences of changes in temperature include sea level rise, extreme weather events, and disruptions to ecosystems and agriculture.

4. Can we reverse the effects of changes in temperature?

While it may not be possible to completely reverse the effects of changes in temperature, taking actions such as reducing greenhouse gas emissions and promoting sustainable practices can help mitigate their impacts.

5. How can we accurately measure changes in temperature?

Scientists use a variety of tools and methods to measure changes in temperature, including thermometers, satellites, and climate models. These measurements are then compared to historical data to track trends and patterns over time.

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