Question

Specific heat of Al = $0.215\,cal{{\,}^{\circ }}{{C}^{-1}}{{g}^{-1}}$ . 25 gm of Al having ${{T}_{1}}={{90}^{\circ }}C$ is placed in 100 gm of water having $T={{20}^{\circ }}C$ . What is the final temperature?

Answer 1

There is a transfer of energy in the form of heat, until both substances acquire equilibrium at a particular temperature, where the amount of heat required to reach the temperature of unit mass of substance by a degree is given by its specific heat.

Complete step by step answer:
In the given system, as the aluminium is placed in the beaker containing water, there is a flow of thermal energy in the form of heat as both the substances are present in two different temperatures. Such that the heat flows from the higher to the lower temperature until both of them acquire equilibrium.
This heat flow, Q is expressed as follows:
$Q=m\times {{c}_{p}}\times \Delta T$ , where m is the mass of substance, ${{c}_{p}}$ is the Specific heat capacity and $\Delta T$ is the difference in temperature. Since between the two substances involving heat flow, one substance gains the energy, the other losses the energy in equal amounts. Thus, we have,
$\implies\text{Heat}\,\text{lost}\,\text{by}\,\text{Aluminium =}-\text{Heat}\,\text{gained}\,\text{by}\,\text{water}$
$\implies {{m}_{Al}}\times {{c}_{p}}\times ({{T}_{f}}-{{T}_{i}})=-{{m}_{water}}\times {{c}_{p}}\times ({{T}_{f}}-{{T}_{i}})$
-Here ${{T}_{f}}$ is the final temperature at which equilibrium is attained by both the substances. Given the mass of aluminium = 25 g, the specific heat = $0.215\,cal{{\,}^{\circ }}{{C}^{-1}}{{g}^{-1}}$ and initial temperature is ${{T}_{i}}={{90}^{\circ }}C$. As for the water, the mass is 100 g, specific heat is $1.0\,\,cal{{\,}^{\circ }}{{C}^{-1}}{{g}^{-1}}$ and initial temperature is $T={{20}^{\circ }}C$. Then, substituting these values in above relation, we get,
$\implies 25\times 0.215\times ({{T}_{f}}-90)=100\times 1\times ({{T}_{f}}-20)$
$\implies 5.375\times ({{T}_{f}}-90)=-100\times ({{T}_{f}}-20)$
$\implies (100+5.375){{T}_{f}}=2000-483.75$
$\implies 105.375\,{{T}_{f}}=1516.25$
$\implies {{T}_{f}}=14.389\approx {{14.4}^{\circ }}C$

Therefore, the final temperature obtained is ${{14.4}^{\circ }}C$.

Note: The heat energy is actually obtained from the specific heat, which accounts for the amount of heat required to raise the temperature of unit mass of substance by $1K$. The SI unit is Joules/ Kelvin/kilogram. It can be expressed in various other units of energy (calories), mass (gram) and temperature (Celsius and Fahrenheit).