Physics_Phailure Posts: 1, Reputation: 1 New Member #1 Apr 3, 2007, 06:25 PM
Heat Transfer - Physics
a .500kg sample of water is at 15 degrees C in a calorimeter. A .0400kg mass of zinc at 115 degrees C is placed in the water. What is the final temp of the system if specific heat of zinc is 388J/kg/C and the specific heat of water is 4184J/kg/C?

This was the exact question goven to me I think it has something to do with the formula Q=mcDeltaT and energy has to always be at a constant since it can't be either destroyed or created, BUT don't take my word on that.

ANY help wrong or hopefully otherwise would be greatly appreciated.
 Capuchin Posts: 5,255, Reputation: 656 Uber Member #2 Apr 3, 2007, 11:29 PM
Well, the water has to gain the same energy as the zinc loses, and the system has to come to an equal temperature.

Using these 2 pieces of information, you should be able to form equations and solve them.
 NITESH KUMAR SAHA Posts: 4, Reputation: 1 New Member #3 Apr 5, 2007, 04:08 AM
Originally Posted by Physics_Phailure
a .500kg sample of water is at 15 degrees C in a calorimeter. a .0400kg mass of zinc at 115 degrees C is placed in the water. what is the final temp of the system if specific heat of zinc is 388J/kg/C and the specific heat of water is 4184J/kg/C?

This was the exact question goven to me i think it has something to do with the formula Q=mcDeltaT and energy has to always be at a constant since it can't be either destroyed or created, BUT dont take my word on that.

ANY help wrong or hopefully otherwise would be greatly appreciated.
m1*s1*[T-(273+15)] = m2*s2*[273+115-T]

(m1*s1+m2*s2)T = m2*s2*388 +m1*s1*288

T= {(.04*388*388)+(.5*4184*288)}/(.04*388 +.5*288)

= X kelvin

=X-273 deg c

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