Are you suggesting that I use 4.184 for the heat capacity of every one? If I do then it wouldn't make sense since that's just making the left side equal to the right side and getting at final temperature of 25.0°C.
By the way, is my formula correct though?
heat of solution = heat of water...
So here's the problem
Conside the dissolution of CaCl2:
CaCl2(s) ---> Ca^(2+) (aq) + 2Cl^(-1) (aq) deltaH = -81.5kJ
An 11.0g sample of CACl2 id dissolved in 125g of water, with both substances at 25.0°C. Calculate the final temperature of the solution assuming no heat lost to...
M=1.99 kg, R=0.133 m, F=8.142 N
I = .5MR^2, T(1) = r x F and T(2) = I x alpha.
Solve for I and then solve T(1) using F given and then once you have both plug into T(2).
I've been at this problem for almost an hour and I know that I'm going in the right direction but I think that I'm using something wrong within the calculations and what not. But well...here's the problem:
A rod of length 61.0 cm and mass 1.10 kg is suspended by two strings which are 44.0 cm...