A metal object of mass 30.6g was heated to 30.0 degrees Celsius and placed into

Question

A metal object of mass 30.6g was heated to 30.0 degrees Celsius and placed into 20.0mL of water at 19.1 degrees Celsius. (density=1.00g/mL). A thermometer was monitored and a temperature of 20.0 degrees Celsius was recorded as the highest temperature for the calorimetric experiment.

a. Calculate the Q for the water in Joules (c for water is 4.184J/g degrees celsius)

b. Determine the Q for the metal in Joules based on your knowledge of conservation of energy. (you cannot use the Q=mcdeltaT equation, because you do not know the c for metal).

c.Based on the Q value found in B, calculate the specific heat of the metal.

d. You instructor informs you that the metal object was made of silver and has a specific heat value of 0.235J/g degrees Celsius. Calculate the percent error.

Explanation / Answer

a. Q=msDelta T

Q= 20*4.184*10 = 836 J

b. Q= 836 j ( According to law of conservation of energy. energy neither created nor destroyed)

c. specific heat of metal s= Q/m*Delta T = 836/30.6*10 = 2.7326 j/gram.0C

d. Q= 30.6*0.235*10 = 71.91 j

actual Q= 836 j

error = 91.39%

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