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A 2 kg copper block, initially at 100°C, is brought in contact with a 2 kg plastic block, initially at 20°C. They are kept in contact long enough to reach the same equilibrium temperature. The specific heat of plastic is four times as much as the specific heat of copper.
Which block will go through a larger temperature change? Why?
Since this is a problem early in the first few chapters of a Heat Transfer class, I assume that the two blocks can be treated as a closed system and no heat is lost to the surroundings.
Here is my working, which arrived at an answer that seems illogical.Energy lost by copper = Energy gained by plastic
Since there is negligible change of other energies such as kinetic, potential, etc, assume that,
Thermal energy lost by copper = Thermal energy gained by plastic
[tex]mc \Delta T_{copper} = mc \Delta T_{plastic}[/tex]
Since the masses of both blocks are equal,
[tex](c \Delta T)_{copper} = (c \Delta T)_{plastic}[/tex]
And since the specific heat of the plastic is four times that of the copper,
[tex]c_{copper} \Delta T_{copper} = 4c_{copper} \Delta T_{plastic}[/tex]
[tex]c_{copper}(T_{final} - 373) = 4c_{copper}(T_{final} - 293)[/tex] (temperatures in Kelvin)
[tex]T_{final} - 373 = 4T_{final} - 1172[/tex]
[tex]T_{final} = 266[/tex]
Therefore, [tex] \Delta T_{copper} = 266 - 373 = -107 K[/tex]
and [tex] \Delta T_{plastic} = 266 - 293 = -27 K[/tex]Since the cooler plastic block must gain heat from the warmer copper block, it is impossible that the temperature change for the plastic block could be negative. What's wrong with my calculations?
Edit: Latex seems to format properly now, thanks Pseudo.
Which block will go through a larger temperature change? Why?
Since this is a problem early in the first few chapters of a Heat Transfer class, I assume that the two blocks can be treated as a closed system and no heat is lost to the surroundings.
Here is my working, which arrived at an answer that seems illogical.Energy lost by copper = Energy gained by plastic
Since there is negligible change of other energies such as kinetic, potential, etc, assume that,
Thermal energy lost by copper = Thermal energy gained by plastic
[tex]mc \Delta T_{copper} = mc \Delta T_{plastic}[/tex]
Since the masses of both blocks are equal,
[tex](c \Delta T)_{copper} = (c \Delta T)_{plastic}[/tex]
And since the specific heat of the plastic is four times that of the copper,
[tex]c_{copper} \Delta T_{copper} = 4c_{copper} \Delta T_{plastic}[/tex]
[tex]c_{copper}(T_{final} - 373) = 4c_{copper}(T_{final} - 293)[/tex] (temperatures in Kelvin)
[tex]T_{final} - 373 = 4T_{final} - 1172[/tex]
[tex]T_{final} = 266[/tex]
Therefore, [tex] \Delta T_{copper} = 266 - 373 = -107 K[/tex]
and [tex] \Delta T_{plastic} = 266 - 293 = -27 K[/tex]Since the cooler plastic block must gain heat from the warmer copper block, it is impossible that the temperature change for the plastic block could be negative. What's wrong with my calculations?
Edit: Latex seems to format properly now, thanks Pseudo.
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