Question

a. Calculate the energy transferred when the temperature of 75 cm3 of water rises from 23 °C to 54 °C.

b. When 8 g of sodium chloride is dissolved in 40 cm3 of water the temperature falls from 22 °C to 20.5 °C. Calculate the energy absorbed by the solution when sodium chloride dissolves.

c. A student added 50 cm3 of sodium hydroxide to 50 cm3 of hydrochloric acid. Both solutions were at 18 °C to start with. When the solutions were mixed a reaction occurred. The temperature rose to 33 °C. Calculate the energy released in this reaction.

 

Answer 1

Ans A) 

Given,  

( Temperature ) T1 = 23°C 

T2 = 54° C  

 75g of water= 75cm^3 of water 

q =? 

We know that 

 q = mc∆T 

Here = ∆T = T2-T1

 q = 75 x 4.186 x (54- 23)

q= 9732.45J

Hence , energy transferred = 9.7324 KJ 

  

Ans B) Given , 

Mass ( m) = 8g 

 T1 = 22° C 

T2 = 20.5 °C  

mass of 40cm3 = 40g  

According to the formula 

 q = mc∆T

  q = 40 x 4.186 x (20.5- 22)

q = -251.16J

 

Hence , energy absorbed = 251.16J for 8g of NaOH 

And for 1mole NaOH, heat absorbed = 1255.8J/mol

 

Ans C) Given , 

 T1 = 18 ° C 

T2 = 33° C  

 V1 { V Of NaOH} = 50cm^ 3 

V2 { V of HCl} = 50cm^3 

Total Volume = 50cm^3 + 50cm^ 3 = 100cm^3 

V = 100cm^3 

 Here , we assume the density = 1g/cm^3  

mass = 100g

Let the specific heat be equal to the water

So, q = mc∆T

 q = 100 x 4.186 x (33- 18)

q = 6279J

Hence , energy released = 6.279KJ.

So, q = mc∆T

 q = 100 x 4.186 x (33- 18)

q = 6279J

Hence , energy released = 6.279KJ.