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

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

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explanation: