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How much heat is absorbed when a 298.3 g piece of brass goes from 30.0 to150 degrees Celsius? The specific heat of brass is .038 J/goC. PLEASE HELP

Sagot :

Neetoo

Answer:  

Heat absorbed = 1360.248 j  

Explanation:  

Given data:  

Mass of brass = 298.3 g  

Initial temperature = 30.0°C  

Final temperature = 150°C  

Specific heat capacity of brass = 0.038 J/g.°C  

Heat absorbed = ?  

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:  

Q = m.c. ΔT  

Q = amount of heat absorbed or released  

m = mass of given substance  

c = specific heat capacity of substance  

ΔT = change in temperature  

ΔT = 150°C - 30.0°C    

ΔT = 120°C  

Q = 298.3 g × 0.038 J/g.°C × 120°C  

Q = 1360.248 j

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