Q1 of 11 Page 89

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.)

Given

Mass of Ice at -10°C = 2Kg


To Find – Quantity of heat required to completely melt the ice.


Formula


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 ice (J/kg-K)


ΔT is the temperature difference(K)


Step 1


Heat required to change temperature of ice from -10°C to 0°C.


ΔT = Initial temperature – Final temperature


= 263K – 273K


= -10K


Q = mc(ΔT)


m = 2kg


c = 2.1 × 103 J/kg K


ΔT = -10K


Q1 = 2Kg × 2.1 × 103 J/kg K × (-10K)


Q1 = - 42 × 103 J (Negative sign shows that the heat is absorbed)


Step 2


Heat required to convert ice to water at 273K.


Q = mF


m = 2kg


F = 336 × 103 J/kg


Q2 = 2kg × 336 × 103 J/kg


Q2 = 672 × 103 J


Total heat required is Q1 + Q2 = 714 × 103 J


More from this chapter

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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.

 7

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

 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

 3

A hole was drilled in an ice block at 273 K. When the hole was filled with water at 373 K, 2 kg of ice melted and changed into water at 273 K. If so, what is the mass of water at 373 K that was used? (Specific heat capacity of water 4200 J kg–1 K–1, latent heat of fusion of ice: 336 × 103 J/kg)