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Do some converting first. 688 calories (4.184 Joules/1 calorie) = 2878.592 Joules 25 ml of water = 25 grams q(Jolules) = mass * specific heat * (Temp. final - Temp. initial) 2878.592 Joules = 25 grams Water * 4.180 J/gC * (Temp Final - 80C ) 2878.592 Joiles = 104.5( Temp. Final) - 8360 11238.592 =104.5(Temp. Final) 107.55 Celsius Final Temperature ( call it 108 C )
1.77
The temperature of the metal bar decreases.The temperature of the cool water increases.The final temperature of the metal bar will be the same as the final temperature of the water.
The final temperature will depend not only on the initial and final pressures, but also on the initial temperature and whether the expansion is adiabatic.
To calculate the delta temperature, you will take the difference between the final and initial temperature.
You need to know the specific heat for copper. Then use q = mc∆T for copper and water. Heat lost by the copper MUST equal heat gained by the water. You can then solve for T2 of the water.
This would depend on the temperature of the water before you add the metal and what type of metal it is. if its copper it absorbs heat fast and would not change the temperature much but if you dropped lead into it then it would have to absorb more heat making the temperature lower than the copper. There are too many variables to answer the question.
If you can measure 3 of these 4 things then you can use this formula q( energy in Joules ) = Mass * specific heat * temperature final - temperature initial
Final Temperature Initial Temperature Specific Heat Capacity of Calorimeter Plug the values into the equation: q = C( Tf - Ti ) , where C = specific heat, Tf = Final Temperature, and Ti = Initial Temperature.
Copper is refined in a smelter. Then once raw copper is obtained the final purification is via electroplating.
Q=m.c.Δt (m: mass, c: specific heat capacity, Δt :change in temperature)(specific heat capacity of aliminum is 0.90 J/K.g = 200 x 0.90 x 10 (if initial temperature is 0)(if not, then Δt=final temperature-initial temperature) =1800 Joules
When allowed to stand for long enough, the final temperature will reach room temperature.
When allowed to stand for long enough, the final temperature will reach room temperature.
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