How Do I Solve These Chem Problems? (Physical Thermochemistry) ?
My chemistry teacher was not good during distance learning at the end of last year so I have no idea what to do for these problems in my AP chem summer work.
1. A 15.3g piece of metal with an initial temp of 100.9°C is added to 90.3g of water with an initial temp of 20.9°C. The final temp is 27.0°C. Calculate the specific heat of the metal.
2. A 10.4g piece of metal with an initial temperature of 100°C is added to 33.3g of water with an initial temp of 22.0°C. The specific heat of the metal is 0.593J/g°C. Calculate the final temp. (Cp for water is 4.18 J/g°C)
3. What is the enthalpy of solution in kJ/mole if 5.05g of NaOH is dissolved into 124g of water? The temp of the water changes from 23.1°C to 26.4°C. Is this exothermic or endothermic? Explain.
4. When ice at 0°C melts to liquid at 0°C, it absorbs 0.334 kJ of heat per gram. Suppose the heat needed to melt 38.0 g of ice is absorbed from the water contained in a glass. If this water has a mass of 0.210kg and a temperature of 21.0°C, what is the final temp of the water? (Note that you will also have 38.0g of water at 0°C from the ice.)
Roger the Mole2020-08-08T03:35:40Z
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1. First, calculate the amount of heat gained by the water: (4.184 J/g·°C) x (90.3 g) x (27.0 - 20.9)°C = 2304.67 J gained by the water
The same amount of heat was lost by the metal, so: (2304.67 J) / (15.3 g) / (100.9 - 27.0)°C = 2.04 J/g·°C (the requested specific heat)
(Although it is difficult to imagine a metal with a specific heat that high which does not react with water.)
2. Let T be the final temperature (in °C) to be found. (0.593 J/g°C) x (10.4 g) x (100 - T)°C = (616.72 - 6.1672 T) J lost by the metal
(4.184 J/g·°C) x (33.3 g) x (T - 22.0)°C = (139.327 T - 3065.2) J gained by the water
Set to two expressions for the number of joules lost or gained equal to each other: 616.72 - 6.1672 T = 139.327 T - 3065.2 Solve for T algebraically: T = 25.3°C
3. Supposing the specific heat of the solution is the same as for water: (4.184 J/g·°C) x (5.05 g + 124 g) x (26.4 - 23.1)°C = 1781.82 J produced by the dissolution
And since the temperature rose, the dissolution is exothermic.
(1781.82 J) / (5.05 g NaOH / (39.99715 g NaOH/mol)) = 14112 J/mol = -14.1 kJ/mol
(The final minus sign did not come from the calculations. It is the convention for exothermic processes like this one.)
4. First, calculate the amount of heat required to melt the ice: (334 J/g) x (38.0 g) = 12692 J
Let T be the final temperature (in °C) to be found.
(4.184 J/g·°C) x (38.0 g) x (T - 0)°C = (158.992 T) J required to warm the melted ice to T
(4.184 J/g·°C) x (210 g) x (21.0 - T)°C = (18451.44 - 878.64 T) J total lost by the warm water
Set the expressions for heat required or lost equal to each other (units in joules): 12692 + 158.992 T = 18451.44 - 878.64 T Solve for T algebraically: T = 5.55°C