Question

An aluminium vessel of mass 0.5kg contains 0.2kg of water at $20^\circ C$ . A block of iron of mass 0.2kg at $100^\circ C$ is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are ${\text{910 J k}}{{\text{g}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}{\text{, 470 J k}}{{\text{g}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}{\text{, 4200 J k}}{{\text{g}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}$ respectively.

Answer 1

Hint : In thermodynamics, a substance's specific heat capacity, also known as massic heat capacity, is the heat capacity of a sample divided by the mass of the sample. Informally, it is the quantity of heat that must be given to one unit of mass of a substance to induce a temperature increase of one unit. The joule per kelvin per kilogramme is the SI unit for specific heat capacity.

Complete Step By Step Answer:
When the passage of heat causes temperature to rise or fall, this is referred to as a temperature change. This causes chemical equilibrium to move toward the products or reactants, which may be detected by looking at the reaction and determining if it is endothermic or exothermic. A change in temperature, pressure, or concentration of reactants in an equilibrated system will induce a reaction that partially offsets the change to establish a new equilibrium, according to Le Chatelier's principle. Adding or withdrawing heat alters the equilibrium when the temperature is changed. Chemical reactions are frequently formulated in such a way that the passage of heat in the reaction is not explicitly addressed.
From the question
Mass of aluminium $= 0.5\;{\text{kg}}$
Mass of iron $= 0.2\;{\text{kg}}$
Specific heat of iron $= {100^\circ }{\text{C}} = {373^\circ }{\text{K}}$
Specific heat of iron $470\;{\text{J K}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$
Mass of water $= 0.2\;{\text{kg}}$
Temp. of aluminium and water $= {20^\circ }{\text{C}} = {297^\circ }{\text{K}}$
Specific heat of aluminium $= 910\;{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$
Specific heat of water $4200\;{\text{JK}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$
So the
Heat gain $= 0.5 \times 910(\;{\text{T}} - 293) + 0.2 \times 4200 \times (343 - {\text{T}})$
$\therefore ({\text{T}} - 292)(0.5 \times 910 + 0.2 \times 4200)$
also
Heat lost $= 0.2 \times 470 \times (373 - {\text{T}})$
We know that, Heat gain = Heat lost
So we substitute and get
$\therefore ({\text{T}} - 292)(0.5 \times 910 + 0.2 \times 4200) = 0.2 \times 470 \times (373 - {\text{T}})$
$\therefore ({\text{T}} - 293)(455 + 8400) = 49(373 - {\text{T}})$
Hence
$\therefore ({\text{T}} - 293)\left( {\dfrac{{1295}}{{94}}} \right) = (373 - {\text{T}})$
$\therefore ({\text{T}} - 293) \times 14 = 49(373 - {\text{T}})$
So the temperature is $\Rightarrow {\text{T}} = \dfrac{{4474}}{{15}} = 298\;{\text{K}}$
$\therefore {\text{T}} = 298 - 273 = {25^o} - {0^o}{\text{C}}$
Final temperature $= {25^\circ }{\text{C}}$ .

Note :
For gases, liquids, and solids with a reasonably general composition and molecular structure, the specific heat capacity may be defined and measured. Gas mixtures, solutions, and alloys, as well as heterogeneous materials like milk, sand, granite, and concrete, may all be studied on a vast scale. As long as the changes are reversible and gradual, the specific heat capacity may be determined for materials that change state or composition as temperature and pressure vary. For example, the ideas may be defined for a gas or liquid that dissociates as the temperature rises, as long as the dissociation products recombine quickly and fully when the temperature falls.