Question

A sphere of aluminium of mass $ 0.047kg $ placed for sufficient time in a vessel containing boiling water, so that the sphere is at $ {100^o}C $ .It is then immediately transferred to $ 0.14kg $ copper calorimeter containing $ 0.25kg $ of water at $ {20^o}C $ . The temperature of water rises and attains a steady state at $ {23^o}C $ .Calculate the specific h capacity of aluminium. Given : specific heat capacity of water is $ 4186Jk{g^{ - 1}}{K^{ - 1}} $ , specific heat capacity of copper calorimeter is $ 385Jk{g^{ - 1}}{K^{ - 1}} $ .

Answer 1

Hint: This question is based on the concept of specific heat capacity. Specific heat capacity is defined as the amount of heat required to raise the temperature of 1 kilogram of a substance by 1 Kelvin. The SI unit of specific heat capacity is $ Jk{g^{ - 1}}{K^{ - 1}} $ .

Complete answer:
Specific heat capacity depends upon the following factors:
(A) Nature of substance
(B) Temperature
Given,
Mass of the aluminium sphere \[{m_1} = 0.047kg\]
Initial temperature of aluminium sphere $ {100^o}C $ $ $ $ $
Final temperature of aluminium sphere $ {23^o}C $
Change in temperature $ (\Delta {T_1}) = {100^o}C - {23^o}C $
 $ (\Delta {T_1}) = {77^o}C $
Now, let the specific heat capacity of aluminium be $ {c_1} $
The amount of heat which is lost by the aluminium $ = {m_1} \times {c_1} = \Delta {T_1} $
 $ = .047kg \times {c_1} \times {77^o}C $
Now, we are given that the mass of the water $ {m_2} = 0.25kg $
Also, the mass of calorimeter $ {m_3} = 0.14kg $
The initial temperature of water and calorimeter $ = {20^o}C $
The final temperature $ = {23^o}C $
Change in temperature $ (\Delta {T_2}) = {23^o}C - {20^o}C $
 $ (\Delta {T_2}) = {3^o}C $
Also, the specific heat of water $ ({c_2}) = 4.18 \times {10^3}Jk{g^{ - 1}}{C^{ - 1}} $
Now, the total amount of energy gained by the water and calorimeter,
 $ {m_2} \times {c_2} \times \Delta {T_2} + {m_3} \times {c_3} \times \Delta {T_2} $
On putting the required values, we get,
 $ = 0.25 \times 4.18 \times {10^3} \times 3 + 0.14 \times 0.386 \times {10^3} \times 3 $
 $ = 3297.12J $
Since this is steady state,
Heat lost by aluminium = heat gained by water + heat gained by calorimeter
Thus, we can say that,
 $ 0.047kg \times {c_1} \times {77^o}C = 3297.12J $
On further solving this, we get,
 $ {c_1} = 911.058Jk{g^{ - 1}}{C^{ - 1}} $
So, the specific heat capacity of aluminium is $ {c_1} = 911.058Jk{g^{ - 1}}{C^{ - 1}} $.

Note:
Specific heat capacity is an intensive property. An intensive property is one that does not depend on the mass of the substance or system. No matter how much aluminium we have, its ability to absorb heat energy at a particular temperature is the same.