Q3 of 11 Page 89

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)

Given

Mass of water at 373K added = ?

Mass of ice melted and converted into water = 2Kg

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

Energy absorbed in conversion of ice to water at 273K

Q = mF

F = 336 × 103 J/kg

M = 2Kg

Q = 672 × 103 J


Step 2

Energy released in changing the temperature of ‘m’ kg water from 373K to 273K .

Q = mc (ΔT)

M = ?

C = 4200 J kg–1 K–1

ΔT = 100K

Q = m(kg) × 100K × 4200 J kg–1 K–1

Since no e× ternal heat/ energy is provided , the heat absorbed in Step 1 is equal to heat released in Step 2.

Equating

m(kg) × 100K × 4200 J kg–1 K–1 = = 672 × 103 J

We get m = 1.6Kg

Mass of water added equal to 1.6 Kg.

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7

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

 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

 4

Collect information regarding different kinds of thermometers and record the situations in which they are used. (hint: ma× imum – minimum thermometer).