A 130 g sample of brass at 120.0 degrees Celsius is placed in a calorimeter cup that contains 130 g of water at 35.0 degrees Celsius. Disregard the absorption of heat by the cup and
calculate the final temperature of the brass and water. Specific heat of water = 4.18 J/gC,
specific heat of brass=0.380 J/gC. Attach your complete solution

Explanation:

Hello there!

In this case, since we are looking for an equilibrium temperature, we can evidence how the net heat flow is 0 as all the heat released by the hot brass is absorbed by the water:

Thus, by writing that equation in terms of mass, specific heat and temperature we obtain:

Thus, by applying the following algebra, it is possible to arrive to a clean expression to obtain the equilibrium temperature:

Then, by plugging the masses, specific heats and initial temperatures in, we obtain:

Best regards!

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

16

explanation:

the marks go by 4s