1. A. Define specific heat B. Why do you have to transfer the copper rapidly fro
ID: 1043440 • Letter: 1
Question
1. A. Define specific heat B. Why do you have to transfer the copper rapidly from the boiling water into the calorimeter? C. what error would be committed if the boiling water did not cover the metal completely? 2. A student places 1.38 g of unknown metal at 99.6°C into 60.50 g of water at 221°C. The entire system reaches a uniform temperature at 31.6 C. Calculate the specific heat of the metal 3. If the actual specific heat of the metal in Problem 2 is 0.25 J/g. °C, calculate the percentage error.Explanation / Answer
Solution:- 1A:- Heat required to raise the temperature one gram of a substance by one degree C is known as it's specific heat.
For example, Specific heat of water is 4.184 J/g.0C. It merans 4.184 J of heat is required to raise 1 gram water temperature by 1 degree C.
B) We need to transfer the hot copper rapidly from boiling water to calorimeter so that the heat is not lost to the surroundings.
C) If the boiling water does not cover the metal then metal's temperature might not be uniform and also it would be losing the heat to the surroundings. This may affect the temperature of the metal that we recod.
2) Heat given = -heat taken
Water temperature is increasing it means the heat is gained by water.
Q = m c delta T
for water, delta T = 31.6 - 22.1 = 9.5 degree C
m = 60.50 g
c = 4.184 J/g.degree C
Q = 60.50 g x 4.184 J/g.degree C x 9.5 degree C
Q = 2404.754 J
Equal amount of heat is lost by the metal.
For metal, Q = -2404.754 J
delta T = 31.6 - 99.6 = -68 degree C
m = 1.38 g
-2404.754J = 1.38 g x c x (-68 degree C)
c = 2404.754 J/(1.38g x 68 degree C)
c = 25.6 J/g.degree C
So, the exerimental specific heat of the unknown is 25.6 J/g.degree C.
3) percent error = ([accepted - experimental]/accepted)*100
percent error = ([0.25 - 25]/0.25]*100
= -9900%
It's quit high that means the given data is not good.
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