A 50 grams of metal at 95 degree celcius is dropped into 250 grams of water at 17 degree celcius and warms it up to 19.4 degree celcius. What is the specific heat(c) of the metal?
2007-10-07 21:00:36 · 3 answers · asked by yien 1 in Science & Mathematics ➔ Physics
the result shoul be:0.16 cal/g "c
2007-10-07 21:44:40 · update #1
Q1 = Q2
50 x c1 x ( 95 - 19.4 ) = 250 x 1 x ( 19.4 - 17 )
( assuming that specific heat of the water is 1 )
you can get c1 from this equation
2007-10-07 21:16:08 · answer #1 · answered by Nerd 1 · 0⤊ 0⤋
250g*(19.4-17)=600 cal put into the water. It came from the metal whose temperature dropped by (95-19.4)=75.6 degrees so the specific heat of the metal is 600/(75.6*50) = .158 cal/g-degree.
Doug
2007-10-07 21:18:48 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
Formula: Q = mc(t2-t1)
m: mass; c: specific heat
t: temp.
Eq.
m1c1(95-19,4) = m2c2(19,4-17)
1: for metal
2 for water
c2 = 4.186 joule/gram °C
calculate
c1 = 0.664 joule/gram °C
2007-10-07 21:22:24 · answer #3 · answered by Tung 2 · 0⤊ 0⤋