How much energy was required to heat a 5.00g sample of liquid water at 25.0°C to 100°C? (Specific heat of water is 4.184J/g°C).

Q = 1569 j

Explanation:

Given data:

Mass of water = 5.00 g

Initial temperature = 25.0°C

Final temperature = 100°C

Cp = 4.184 j/g.°C

Energy needed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  100°C - 25°C

ΔT =  75°C

Q = 5 g× 4.184 j/g.°C×75°C

Q = 1569 j

"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"