110 grams of boiling water are poured into an aluminum pan whose mass is 1000 grams. The initial temperature of the water is 100 C, with a heat capacity of 4.2 J/(gram · K). The aluminum pan has an initial temperature of 24 C with a heat capacity of 0.9 J/(gram · K). After a while, the water and pan come to a common temperature. What is that temperature? Approximate the water and pan as a closed system.
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Temp. of Water= 273 + 100 = 373 K
Energy Contained in Water = mcT = mass x specific heat capacity of Water x temperature
= 110 (g) x 4.2 (J/g/K) x 373 (K) = 172326 J
Temp. of Aluminium = 273 + 24 = 297 K
Energy Contained in Aluminium = mcT = mass x specific heat capacity of Aluminium x temperature
= 1000 (g) x 0.9 (J/g/K) x 297 = 267300 J
Total Energy of System = 172326 J + 267300 J = 439626 J
therefore because the end temperature is shared:
(mcT of water as above) + (mcT of aluminium as above) = Total Energy in system
T(mc of water + mc of aluminium) = Total energy
T = Total Energy / (mc of water + mc of aluminium)
T = 439626 / ([110x4.2] + [1000 x 0.9]) = 322.78 K (or) 49.78 C
For a check: plug the numbers into the original equations (mCT for each) and the total energy should be the same.
Energy Contained in Water = mcT = mass x specific heat capacity of Water x temperature
= 110 (g) x 4.2 (J/g/K) x 373 (K) = 172326 J
Temp. of Aluminium = 273 + 24 = 297 K
Energy Contained in Aluminium = mcT = mass x specific heat capacity of Aluminium x temperature
= 1000 (g) x 0.9 (J/g/K) x 297 = 267300 J
Total Energy of System = 172326 J + 267300 J = 439626 J
therefore because the end temperature is shared:
(mcT of water as above) + (mcT of aluminium as above) = Total Energy in system
T(mc of water + mc of aluminium) = Total energy
T = Total Energy / (mc of water + mc of aluminium)
T = 439626 / ([110x4.2] + [1000 x 0.9]) = 322.78 K (or) 49.78 C
For a check: plug the numbers into the original equations (mCT for each) and the total energy should be the same.