A 75-g piece of metal is heated in boiling water to 99.8C and then dropped into water in an insulated beaker. THere is 325 ml of water (density = 1.00g/ml) in the beaker, and its temperature before the metal is added was 15C. The final temperature of the metal and water is 25.1C. What is the specific heat of the metal?
2006-12-08 10:49:51 · 3 answers · asked by bangor j 1 in Science & Mathematics ➔ Chemistry
The heat distribution changes, the total heat referred to 15C (mass*specific heat*degC) does not, so (99.8-15)*75*x = (25.1-15)*(325*4.1855 + 75*x), where x is the specific heat of the metal. Juggling the terms, we have 84.8*75*x = 10.1*325*4.1855 + 10.1*75*x, or x = 10.1*325*4.1855/(74.7*75) = 2.452. (Units are J/(g-degC)).
See the worked example in the ref.
2006-12-08 11:24:40 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
you're able to desire to establish an equilibrium, because of the fact the skill of the water and the metallic might desire to stay consistent. In different words, the skill lost interior the metallic has to equivalent the skill transferred to the water, and the internet skill is 0. frequently, the skill utilized in heating/cooling may well be defined as: (T2-T1)*particular warmth*mass Water has a particular warmth of a million calorie/gram/diploma Celsius, and we will permit x represent the particular warmth of the metallic. because of the fact the internet skill substitute must be 0 energy: (35-20)*a million*one hundred + (35-a hundred and twenty)*x*one hundred fifty = 0 fixing for x, the specifc warmth of the metallic is 0.11765 energy/gram/diploma Celsius
2016-12-11 05:12:18 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Eh = m x c x change in temp (T)
I can say the mass is 0.325 Kg
the T is 10.1 degrees
But i can't figure out what the heat enrgy is before the metal is added, so you can rearrange the equation.
Sorry!
2006-12-08 11:15:20 · answer #3 · answered by Anonymous · 0⤊ 0⤋