# A Cube of Iron (Density = 8000 Kg M−3, Specific Heat Capacity = 470 J Kg−1 K−1) is Heated to a High Temperature and is Placed on a Large Block of Ice at 0°C. - Physics

Sum

A cube of iron (density = 8000 kg m−3, specific heat capacity = 470 J kg−1 K−1) is heated to a high temperature and is placed on a large block of ice at 0°C. The cube melts the ice below it, displaces the water and sinks. In the final equilibrium position, its upper surface just goes inside the ice. Calculate the initial temperature of the cube. Neglect any loss of heat outside the ice and the cube. The density of ice = 900 kg m−3 and the latent heat of fusion of ice = 3.36 × 105 J kg−1.

#### Solution

Given:-

Density of the iron cube = 8000 kg m-3

Density of the ice cube = 900 kg m-3

Specific heat capacity, S = 470 J kg−1 K−1

Latent heat of fusion of ice, L = 3.36 × 105 J kg−1

Let the volume of the cube be V.

Volume of water displaced = V

Mass of cube, m = 8000 V kg

Mass of the ice melted, M = 900 V

Let the initial temperature of the iron cube be T K.

Then,

Heat gained by the ice = Heat lost by the iron cube

m × S × (T − 273) = M × L

⇒ 8000 V × 470 × (T − 273) = 900 V×( 3.36 × 105)

⇒ 376 × 104 × (T − 273) = 3024 × 105

rArrT=(30240+102648)/376

rArrT=132888/376K

rArrT=353.425Kapprox353K

rArrT=353K-273K

rArrT=80^oC

Concept: Measurement of Temperature
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 3 Calorimetry
Q 6 | Page 47