A sample of water drops from an initial temperature of 67.5 C to a final temperature of 25.2 C. If -4425 J of heat were released from this sample of water, what is the mass of the sample?

mass = 25.03 g

Explanation:

Given data:

Initial temperature = 67.5°C

Final temperature = 25.2°C

Heat released = -4425 j

Mass of water = ?

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 = 25.2°C - 67.5°C

ΔT = -42.3°C

Specific heat of water = 4.18 j/g.°C

Q = m.c. ΔT

-4425 j = m × 4.18 j/g.°C × -42.3°C

-4425 j = m × -176.81 j/g

m = -4425 j /  -176.81 j/g

m = 25.03 g

po-215 i got it right on the quiz

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.