Q7 of 11 Page 88

2 kg of water at 293 K is converted completely into ice at 273 K. Calculate the heat liberated.

Given:

Temperature = 293 K


Mass of water = 2kg


To convert water in liquid form to solid form.


Formula used:


Q = mc(ΔT)


Q = mF


where m is the mass (Kg)


F is the latent heat of fusion of ice (J/kg)


c is the specific heat of water (J/kg-K)


ΔT is the temperature difference(K)


Step 1


Heat liberated in conversion from water at 293K to water 273K


ΔT = Initial temperature (of water) – Final temperature (of ice)


= 293K – 273K


= 20K


Q = mc(ΔT)


⇒ Q1 = 2 kg × 4.2×103 J/kg-K × (20K)


= 168 × 103 J


STEP 2


Heat liberated in conversion of water from liquid state to solid state.


Q = mF


⇒ Q2 = 2 Kg × 336 × 103 J/kg


= 672 × 103 J


The water diagram is shown here:



Total heat released = Sum of Q1 + Q2 = 810 × 103 J


Ans = 810 × 103 J


More from this chapter

All 11 →
5

Write down an e× ample to know how the following factors influence the rate of evaporation.

(a) Surface area


(b) Wind

 6

When will cool drinks get cooled faster – on placing ice cubes at 0°C or on adding water at 0°C? Justify your answer.

 1

2 kg of ice at –10°C is continuously heated to melt it completely. Calculate the quantity of heat required. (The latent heat of fusion of ice: 336 × 103 J/kg. The latent heat of vaporization of water: 226 × 104 J/kg. Specific heat capacity if ice: 2.1 × 103 J/kg K. Specific heat capacity of water: 4.2 × 103 J/kg K.)

 2

A solid of initial temperature 0°C is heated. A graph showing changes in the temperature according to the heat supplied is given below.


Based on the graph calculate the following. (Specific heat capacity of the substance: 500J kg–1 K–1)


(a) the mass of solid substance


(b) the latent heat of fusion of the substance