9 A 75 gram sample of water is initially at 30 degrees Celsius. What would it's temperature be if
it absorbed 2000 J of heat?

T2 = 36.38°C

Explanation:

Given data:

Mass of water = 75 g

Initial temperature = 30 °C

Final temperature = ?

Heat absorbed = 2000 J

Solution:

Specific heat capacity of water is 4.18 J/g.°C

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 = Final temperature - initial temperature

2000 J = 75 g×4.18 J/g.°C × T2- T1

2000 J = 313.5 J/°C × T2- T1

2000 J = 313.5 J/°C × T2 - 30 °C

2000 J / 313.5 J/°C  = T2 - 30 °C

6.38 °C = T2 - 30 °C

T2 = 6.38 °C +  30°C

T2 = 36.38°C

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.