Question

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.

Answer 1

Given:

Density of iron $\rho _{Fe} = 8000$ $\frac{kg}{m^{3} }$

Density of ice $\rho_{ice} = 900$ $\frac{kg}{m^{3} }$

Specific heat of iron $S_{Fe } = 470$ $\frac{J}{kg K}$

Latent heat of fusion of ice $L = 3.36 \times 10^{5}$ $\frac{J}{kg}$

To Find: initial temperature of cube,

Let volume displace by iron cube is V

So Mass of cube $m = 8000 V$ kg

Mass of ice $M = 900V$ kg

Let, initial temperature of cube is $T$

⇒ Heat gain by ice = Heat loss by cube

   $ML = mS_{Fe} (T-273)$

$900V \times 3.36 \times 10^{5} = 8000V \times 470 \times (T-273)$

$3024 \times 10^{5} = 376 \times 10^{4} (T-273)$

$T = \frac{132888}{376}$

$T = 353.4$ K

Therefore, the initial temperature of cube is 353.4 K

Answer 2

Answer:

353.4 K

Explanation:

Given:

Density of iron  

Density of ice  

Specific heat of iron  

Latent heat of fusion of ice  

To Find: initial temperature of cube,

Let volume displace by iron cube is V

So Mass of cube  kg

Mass of ice  kg

Let, initial temperature of cube is

⇒ Heat gain by ice = Heat loss by cube

  

K

Therefore, the initial temperature of cube is 353.4 K