Suppose the thermometer used in procedure described in this experiment was in er

Question

Suppose the thermometer used in procedure described in this experiment was in error-all readings were just 1.0 degree low What effect would this have on the calculate specific heat of the metal would it too high too low? or have no effect?Explain. 2. Why is it better to use just one thermometer to read all the temperatures rather than to have one for the metal and another for the water in the calorimeter? 3. Use the Law of Dulong and Petit to estimate the specific heat (in J/gdegreeC) of aluminum. Show calculations. 4. Using this estimated value for the specific of aluminum calculate how many grams of the metal should be taken for this experiment. Assume a) that the metal is heated to 98.0degreeC b) the calorimeter contains 20.0g of water to rise by 10.0 degrees shpw calculation.5. Now repeat Questions 3 and 4, but substitute iron for aluminum-

Explanation / Answer

The amount of heat energy that is required to raise the temperature of one gram of a substance by one degree Celsius is called the specific heat capacity, or simply the specific heat, of that substance.

Since the experiment is not given here I am taking an experiment reported online

The amount of heat energy that is transferred, q, in the process of producing a temperature change can be calculated according to the following equation:

q = s X m X T; s is specific heat, m = mass of the material and T = change in temperature.

A heated sample of this metal will be poured into a calorimeter, consisting of cool water contained in a plastic foam cup. Shortly after mixing, the water and the metal will have come to the same temperature. Because plastic foam is a good insulator, heat cannot easily escape from the calorimeter to the surroundings. Therefore, the heat lost by the metal can be said to be equal to the heat gained by the water.

The amount of heat energy gained by the water will be calculated in the following manner:

The specific heat of water is known. The temperature changes of the water, and of the metal, can be measured, as can the mass of the water and the mass of the metal. Using this data, the specific heat of the metal can be calculated

1) Now as you can see here we are always measuring T the change in temperature so if the thermometer has an error that error is nullified in this experiment and it will not have any effect on the measurementof specific heat of a metal.

2) As you can see the heat lost by the metal is the heat gained by water. If there is any error difference between the two thermometrs used that will have an effect on the measuremenst and hence the calculated specific heat of the metal. So it is better to use one thermometer for the measurement to nullify any erreors.

3)As per Dulong and Petit's law

cM = K

where c is the specific heat, M the atomic weights, and K is 3R or about 25 joules per kelvin(or oC)

atomic weight of aluminium is 26.98g/mol

c = K/M

c = 25 J/oC/26.98g

c = 0.926 J/oC.g

4) As explained earlier the way the experiment is conducted

specific heat (water) X mass (water) X T (water) = specific heat (metal) X mass (metal) X T (metal)

4.179 J/goC x 20 g x 10 = 0.926 J/goC x mass (metal) x (98-T1)

835.8 = 0.926 J/goC x mass (metal) x (98-T1)

Since at the end of the experiment the water and metal will compe to the same temperature and final temperature of the water is 23.0oC + 10 oC = 33.0oC we can take the final temperature of the metal to be 33.0oC.

835.8 = 0.926 J/goC x mass (metal) x (98-33)

mass (metal) = 835.8/0.926 J/goC x 65

mass (metal) = 13.88 g

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