Problem 14.46 A laboratory technician drops an 84.0 g solid sample of unknown ma

Question

Problem 14.46 A laboratory technician drops an 84.0g solid sample of unknown material at a temperature of 100?C into a calorimeter. The calorimeter can is made of 0.145kg of copper and contains 0.210kg of water, and both the can and water are initially at 21.0?C. The final temperature of the system is measured to be 27.0?C. Part A Compute the specific heat capacity of the sample. (Assume no heat loss to the surroundings.) c =   J/(kg?K)  
Problem 14.46 A laboratory technician drops an 84.0g solid sample of unknown material at a temperature of 100?C into a calorimeter. The calorimeter can is made of 0.145kg of copper and contains 0.210kg of water, and both the can and water are initially at 21.0?C. The final temperature of the system is measured to be 27.0?C. Part A Compute the specific heat capacity of the sample. (Assume no heat loss to the surroundings.) c =   J/(kg?K)  
Problem 14.46 Problem 14.46 A laboratory technician drops an 84.0g solid sample of unknown material at a temperature of 100?C into a calorimeter. The calorimeter can is made of 0.145kg of copper and contains 0.210kg of water, and both the can and water are initially at 21.0?C. The final temperature of the system is measured to be 27.0?C. A laboratory technician drops an 84.0g solid sample of unknown material at a temperature of 100?C into a calorimeter. The calorimeter can is made of 0.145kg of copper and contains 0.210kg of water, and both the can and water are initially at 21.0?C. The final temperature of the system is measured to be 27.0?C. A laboratory technician drops an 84.0g solid sample of unknown material at a temperature of 100?C into a calorimeter. The calorimeter can is made of 0.145kg of copper and contains 0.210kg of water, and both the can and water are initially at 21.0?C. The final temperature of the system is measured to be 27.0?C. Part A Compute the specific heat capacity of the sample. (Assume no heat loss to the surroundings.) c =   J/(kg?K)  
Part A Compute the specific heat capacity of the sample. (Assume no heat loss to the surroundings.) c =   J/(kg?K)  
Part A Compute the specific heat capacity of the sample. (Assume no heat loss to the surroundings.) c =   J/(kg?K)  
c =   J/(kg?K)  
c =   J/(kg?K)  
c =   J/(kg?K)  

Explanation / Answer

HEAT LOST BY THE MATERIAL = HEAT GAINED BY WATER + CALORIMETER

M1 S1 ( T1-T) = M2 S2 (T - T2 ) + M3 S3 ( T- T2 )

0.084x S (100 - 27) = 0.210 x 4186 x ( 27 - 21) + 0.145 x 386 x (27 - 21 )

solving we get S = 914.902J/kg.K

all 1 ,2 ,3 subscript represent the material , water and calorimeter parameters respectively.

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