A48 kg sample of water absorbs 366 kj of heat. the water was initially at 20.3 ∘c, what is its final temperature?

The final temperature of water = 22.1224 °C

Explanation:

Step 1 : defien the formula for heat transfer

q = m*c*ΔT

with q = heat transfer

with m = mass of the substance

with c = specific heat ( of water in this case)

with ΔT = difference between final temperature and initial temperature

In this case :

q=366kJ = 366 *10^3 J

mass of water = 48kg = 48000g

c= specific heat of wwater = 4.184 J / g °C

initial temperature = 20.3 °C

final temperature =  TO BE DETERMINED

Step 2: Calculate temperature

366000 J = 48000g * 4.184J/g °C * (T2 - 20.3)

366000 / (480000 * 4.184) = 1.8224 = (T2 - 20.3)

T2 = 20.3 + 1.82 =22.1224 ° C

The final temperature of water = 22.1224 °C

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2