A 0.620 kg PYREX® glass calorimeter contains 0.400 kg of water at 12.8°C (both at thermal equilibrium). At that time, 1,000 kg of solid mercury is added at a temperature of -39°C, which corresponds to the melting point of mercury. a) Describe which parts of the system will give up thermal energy and which will gain thermal energy. thermal energy. Analyze which elements require less energy for phase changes or temperature changes.b) Find the equilibrium temperature. c) Describe the states of water and mercury once thermal equilibrium is reached. cPYREX =703J/Kg*K; cHG =138 J/Kg*K; cH2O =4190 J/Kg*K; cICE =2100 J/Kg*K; LfHG =1,18x104 J/Kg; LfH2O =3,34x105 J/Kg
A 0.620 kg PYREX® glass calorimeter contains 0.400 kg of water at 12.8°C (both at thermal equilibrium). At that time, 1,000 kg of solid mercury is added at a temperature of -39°C, which corresponds to the melting point of mercury. a) Describe which parts of the system will give up thermal energy and which will gain thermal energy. thermal energy. Analyze which elements require less energy for phase changes or temperature changes.b) Find the equilibrium temperature. c) Describe the states of water and mercury once thermal equilibrium is reached. cPYREX =703J/Kg*K; cHG =138 J/Kg*K; cH2O =4190 J/Kg*K; cICE =2100 J/Kg*K; LfHG =1,18x104 J/Kg; LfH2O =3,34x105 J/Kg
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A 0.620 kg PYREX® glass calorimeter contains 0.400 kg of water at 12.8°C (both at
thermal equilibrium). At that time, 1,000 kg of solid mercury is added at a temperature of
-39°C, which corresponds to the melting point of mercury.
a) Describe which parts of the system will give up thermal energy and which will gain thermal energy. thermal energy. Analyze which elements require less energy for phase changes or temperature changes.b) Find the equilibrium temperature.
c) Describe the states of water and mercury once thermal equilibrium is reached.
cPYREX =703J/Kg*K; cHG =138 J/Kg*K; cH2O =4190 J/Kg*K; cICE =2100 J/Kg*K; LfHG =1,18x104 J/Kg; LfH2O =3,34x105 J/Kg
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