A student performed Part A2 and added calcium chloride hut did not stir the mixt
ID: 486117 • Letter: A
Question
A student performed Part A2 and added calcium chloride hut did not stir the mixture at all. Therefore, some solid calcium chloride remained at the bottom of the Styrofoam cup even after he recorded the final temperature of the solution. Would this result in a calculated heat of solution that was higher or lower than the true value? Explain. A student performed Part B and did not allow the metal to heat long enough, so that it did not attain the temperature of the boiling water bath. Would this result in a calculated specific heat that was higher or lower than the true value? Explain. A student performed Pan B and did not wait for the water and metal mixture to reach a temperature maximum. Would this result in a calculated specific heat that was higher or lower than the true value? Explain. A 25.000 g metal sample is cooled from 100.0 degree C to 25.0 degree C. If the heat energy released by the metal is calculated to be 2.00 kJ, what is the specific heat of the metal in J g middot degree C? Show calculations below.Explanation / Answer
Ans.1. CaCl2 releases heat when dissolved in water.
So, when CaCl2 is not completely dissolved (some solid crystal remaining undissolved at bottom), less heat is generated than if it were completely dissolved.
So, the actual increase in the temperature of solution will be lesser than the theoretical (true) increase in temperature.
Ans. 2. Specific heat is the amount of heat required to raise the temperature of 1.0 g substance by 10C.
If the metal is removed before it attains the temperature of 1000C, it will absorb lesser amount of heat. Say, the metal temperature at the time of removal is 950C.
But, the calculations will incorrectly be made assuming that the metal had attained the temperature of 1000C.
So, Actual outcome is : Metal absorbed less amount of heat to reach false presumption of 1000C. Or, the metal absorbs less heat and heats faster.
So, the calculated specific heat of metal will be lower than the theoretical (correct) value.
Ans. 3. If metal and water are not allowed to attain thermal equilibrium, decrease in temperature of metal will be lower, and increase in temperature of water in calorimeter will also be lower than what actually it could have been at thermal equilibrium.
Case 1: Thermal equilibrium no reached in water bath: Calculated specific heat of metal will be lesser than the theoretical value. [see answer 2]
Case 2: Thermal equilibrium no reached in calorimeter: If initial and final temperature are recorded correctly for metal and water in calorimeter, it shall have no effect that the calculations only depend on final and initial values of temperature, irrespective of establishment of thermal equilibrium. For being on safe side and ease of recording measurements, establishment of thermal equilibrium is considered ideal.
Note: It were better if the background information of the experiment was outlined. The principle remains the same but “steps” may differ from text to text. The best way to conclude is to perform calculations under hypothetical “false” measurements. It will give you a much better and clear understanding.
Ans. 4. Total heat lost by metal, q = m x s x dT - equation 1
Where,
q = 2.0 kJ = 2000 J ; [1 kJ = 1000 J]
m = mass in gram = 25.000 g
s = specific heat of water = ?
dT = change in temperature = (final – initial) temperature
= 1000C- 25.00C = 750C
Putting the values in equation 1-
2000 J = 25.000 g x s x 750C
Or, x = 2000 J / (25.000 g x 750C) = 1.066 J / (g0C)
Hence, specific heat of metal = 1.066 J / (g0C) = 1.066 J per gram per 0C.
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