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A 83.2 g sample of metal at 90.31 oC is added to 41.82 g of water that is initially at 19.93 oC. The final temperature of both the water and the metal is 28.42 oC. The specific heat of water is 4.184 J/(goC). Calculate the specific heat of the metal.

Respuesta :

olucheyi

Answer: 0.2885J/g°c

Explanation:

Loss of heat of metal = Gain of heat by the water

Therefore

-Qm = +Qw

Where

Q = mΔTCp

Q = heat

M= mass

ΔT = T.f - Ti

Ti= initial temperature

T.f= final temperature

Cp= Specific heat (m is metal, w is water)

-Qm = +Qw

-[m(T.f-Ti)Cpm] = m(T.f-Ti)Cpw

Therefore

-[83.2g(28.42-90.31)Cpm] = 41.82g(28.42-

19.93)4.194J/g°c

Cpm = 1485.54÷ 5149.25

Cpm = 0.2885J/g°c

Therefore , the specific heat of the metal is 0.2885J/g°c