Question

A mercury thermometer whose markings are in centimetre scale, is used to measure the temperature. It shows 3 cm and 15 cm when kept in melting ice and boiling water respectively. Find the equilibrium temperature, as recorded by the above thermometer when a 10 g iron ball at 229°C is dropped into 50 g of water at 25°C. Here, the water equivalent of the container is negligible and heat lost to the surroundings is almost zero. (Take specific heat capacity of iron and water as $0.1cal{g^{ - 1}}\;^\circ {C^{ - 1}}$and $1cal{g^{ - 1}}\;^\circ {C^{ - 1}}respectively).$

Answer 1

Thermometer is a device used to measure temperature of the system. It is usually filled with mercury in it. When we are provided with the mass of the substances and reading of the thermometer, we have to find its specific heat first. Specific heat is usually defined as a change in heat when the temperature of the system is raised by $1^\circ C$.

Complete step by step answer:
As given,

Mass of iron ball, $\mathop m\nolimits_i = 10g$

Initial temperature, $\mathop t\nolimits_1 = 229^\circ C$

Mass of water, $\mathop m\nolimits_w = 50g$
Initial temperature, $\mathop T\nolimits_1 = 25^\circ C$
We know
Specific heat of water, $\mathop c\nolimits_W = 1cal/g^\circ C$( a point to remember always)

Specific heat of iron, $\mathop c\nolimits_i = 0.1cal/g^\circ C$
Let the equilibrium temperature be T

Heat gained by water, $\mathop \theta \nolimits_1 = \mathop m\nolimits_W \mathop c\nolimits_W \Delta T$

$\mathop \theta \nolimits_1 = 50 \times 1\left( {T - T_1} \right) = 50\left( {T - 25} \right).....................\left( i \right)$Heat lost by iron ball, $\mathop \theta \nolimits_2 = \mathop m\nolimits_i \mathop c\nolimits_i \Delta T$

$\mathop \theta \nolimits_2 = 10 \times 0.1 \times \left( {T_1 - T} \right) = \left( {229 - T} \right).............\left( {ii} \right)$

At equilibrium heat gained by water = heat lost by iron ball, therefore $\mathop \theta \nolimits_1 = \mathop \theta \nolimits_2$
$\Rightarrow 50\left( {T - 25} \right) = \left( {229 - T} \right)$
$\left[ {from{\text{ }}\left( i \right){\text{ }}and{\text{ }}\left( {ii} \right)} \right]$
\Rightarrow 50T - 1250 = 229 - T$$$$ \Rightarrow 50T + T = 1250 + 229Now, $T = \dfrac{{1479}}{{51}}$
$T{\text{ }} = {\text{ }}29^\circ C\;$
When we have provided with reading on thermometer we convert it into ° C by using formula:
$$$$Now, $\mathop l\nolimits_1$= length at given temperature
$\mathop l\nolimits_0$ = length at 0° C and it is given as 3 cm $\mathop l\nolimits_{100}$= length at 100°
$C{\text{ }} = {\text{ }}15{\text{ }}cm$
Temperature in C we have found = $29^\circ {\text{ }}C$
Putting values in equation iii
$29 = \dfrac{{\mathop l\nolimits_1 - 3}}{{15 - 3}} \times 100 = 6.48cm$

Note: A clinical thermometer measures temperature in the range of $35^\circ C{\text{ }} to {\text{ }}45^\circ C$. The standard unit of temperature is kelvin. But when we have to find the marking on the thermometer and we are given with length we use the unit as Celsius.