Indicate whether the error described below will result in a higher, lower or no

Question

Indicate whether the error described below will result in a higher, lower or no change in the calculated specific heat. Explain (3 points each) The balance used to determine the mass of the metal was in error by 10 g on the high side, that is, the recorded mass was larger by 10 g as compared to the true mass. It was thought that the metal was in the boiling water long enough to reach its initial temperature of 100 degree C but it actually wasn't. warm distilled water was used instead of room temperature distilled water in the calorimeter. Twice as much water was used in the calorimeter than specified in the procedure. The initial temperature of the water in the calorimeter was measured and recorded. After 30 minutes the hot metal was added to the calorimeter. Unknown the experimenter, the temperature of the water in the calorimeter before the addition of the metal was 2 degrees warmer

Explanation / Answer

The equation used to determine the specific heat of the metal is

qmetal=qwater=Heat transfer

Massmetalx(Specific heat)metalx(Change in temperature)metal=Masswaterx(Specific heat)waterx(Change in temperature)water

(Specific heat)metal=Masswaterx(Specific heat)waterx(Change in temperature)water/Massmetalx(Change in temperature)metal

A) If the recorded mass is 10 g higher than the true mass

Actual (Specific heat)metal=Masswaterx(Specific heat)waterx(Change in temperature)water/(True Mass)metalx(Change in temperature)metal

So calculated (Specific heat)metal=Masswaterx(Specific heat)waterx(Change in temperature)water/(Recorded Mass)metalx(Change in temperature)metal

=Masswaterx(Specific heat)waterx(Change in temperature)water/(True mass +10 g)metalx(Change in temperature)metal

As the mass of metal is in denominator, any increase in the mass would mean reduction of value of specific heat

as specific heat of metal is inversely proportional to mass of metal. So if the recorded mass of the metal is higher than the true mass, then calculated specific heat of the metal will be lower than the actual specific heat of the metal.

B) If the water wasn't in the boling water to reach its initial temperature of 100oC then that would mean it would have a lower initial temperature than 100oC so actual change in temperature (Tfinal-Tinitial) will also be lesser than the calculated change in temperature. Again

Actual (Specific heat)metal=Masswaterx(Specific heat)waterx(Change in temperature)water/(Mass)metalx(Actual Change in temperature)metal

So calculated (Specific heat)metal=Masswaterx(Specific heat)waterx(Change in temperature)water/(Mass)metalx(Recorded Change in temperature)metal

Since change in temperature is in the denominator, so any increase in temperature would mean a lower specific heat value as Change in temperature of metal and specific heat of metal are inversely related. So a higher recorded change in temperature would mean a lesser recorded specific heat than the actual specific heat.

C) When warm distilled water is used instead of room temperature distilled water it means the actual initial temperature of distilled water is higher than recorded, so actual change in temperature of water is lesser than recorded change in temperature of water.

Actual (Specific heat)metal=Masswaterx(Specific heat)waterx(Actual Change in temperature)water/(Mass)metalx(Actual Change in temperature)metal

So calculated (Specific heat)metal=Masswaterx(Specific heat)waterx(Recorded Change in temperature)water/(Mass)metalx(Change in temperature)metal

Now change in temperature of water is directly proportional to specific heat of metal, so if the recorded change in temperature of water is higher than the actual change in temperature of water, so it will give a higher recorded value of specific heat of metal than the actual value of specific heat of metal.

D) If twice as much water was used in the calorimeter than specified in the procedure than that means that the actual mass of water is more than the recorded mass of the water.

Actual (Specific heat)metal=(Actual Mass)waterx(Specific heat)waterx(Change in temperature)water/(Mass)metalx(Actual Change in temperature)metal

So calculated (Specific heat)metal=(Recorded Mass)waterx(Specific heat)waterxChange in temperature)water/(Mass)metalx(Change in temperature)metal

Also mass of water is directly proportional to the specific heat of metal, so a lower recorded mass of water would mean a lower specific heat of metal than the actual value

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