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# Specific Heat of a metal.?

hi guys, would anybody help me to get the answer.

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- lenpol7Lv 73 years agoFavorite Answer
Energy transferred to water

Q = mcDeltaT

Q = 100g x 4..18 J g^-1K^-1 x(100 - 10)K

Q = 100g x 4.18 J g^-1K^-1 x 90K

Q = 150480J (transferred to the water from the metal).

Energy transferred from the metal

Q = mcDeltaT

c = Q / mDeltaT

c = 150480 / 10g x ( 250 - 20)K

c = 150480 / (10g x 230K )

c = 150480 J / 2300 gK

c = 65.426 J g^-1 K^-1

NB

I have calculated using the modern S.I. nomenclature of units. ; Joules(J) & Kelvin(K)

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