Question

A 2 kg copper block is heated to 500℃ and then it is placed on a large block of ice at 0℃. If the specific heat capacity of copper is 400J/kg/℃ and latent heat of water is $ 3.5 \times 105 $ $ \dfrac {J}{kg} $. The amount of ice that can melt is:
A. $ \dfrac {7}{8} $ kg
B. $ \dfrac {7}{5} $ kg
C. $ \dfrac {8}{7} $ kg
D. $ \dfrac {5}{7} $ kg

Answer 1

Hint:Here we use the basic principle that “ Heat loss by system is heat gained by system“. The below mentioned formulae and terms we use to solve the above questions.

Formula used:
${E_h} = mxl$, where ${E_h}$ = energy, m = mass and l = specific latent heat of fusion for that material.
$Q = mc\Delta T$, where Q = heat added or subtracted, m = mass, c = specific heat and ∆T= change in temperature.

Complete step by step solution:
Given,
m(copper) = 2kg ,
∆T=500℃-0℃=500℃ ,
c = 400 J/kg/℃,
$ l = 3.5 \times 105 $ $ \dfrac {J}{kg} $,
For ice
Heat energy is needed to go into a material to change it from a solid to a melted liquid:
${E_h} = mxl$, where ${E_h}$ = energy, m = mass and l = specific latent heat of fusion for that material.
${E_h}$ = m(ice) $\times$ 3.5 $\times$ 105
For copper
$
Q = mc\Delta T\\
=2 \times 400 \times 500\\
=4 \times 10^5\:
$
The principle of calorimetry indicates the law of conservation energy, i.e. the total heat lost by the hot body is equal to the total heat gained by the cold body.
heat loss =heat gain
$mc\Delta T = ml$
∴m(ice) = $\dfrac{{4 \times {{10}^5}}}{{3.5 \times {{10}^5}}} = \dfrac{8}{7}$
So, the correct answer is “Option C”.

Note: The amount of heat that must be added (or removed) from a unit mass of a substance to change its temperature by one degree Celsius. It is an intensive property.
The specific latent heat of a substance is the amount of energy needed to change the state of 1 kg of the substance without changing its temperature.