sphere of $0.047 \;kg$ aluminium is placed for sufficient time in a vessel containing boiling water, so that the sphere is at $100\,^{\circ} C .$ It is then immediately transfered to $0.14 \;kg$ copper calorimeter containing $0.25\; kg$ water at $20\,^{\circ} C$. The temperature of water rises and attains a steady state at $23\,^{\circ} C$. Calculate the specific heat capacity of aluminium in $kJ\;kg^{-1} K^{-1}$
sphere of $0.047 \;kg$ aluminium is placed for sufficient time in a vessel containing boiling water, so that the sphere is at $100\,^{\circ} C .$ It is then immediately transfered to $0.14 \;kg$ copper calorimeter containing $0.25\; kg$ water at $20\,^{\circ} C$. The temperature of water rises and attains a steady state at $23\,^{\circ} C$. Calculate the specific heat capacity of aluminium in $kJ\;kg^{-1} K^{-1}$
Answer In solving this example, we shall use the fact that at a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter.
Mass of aluminium sphere $\left(m_{1}\right)=0.047 kg$
Initial temperature ofaluminiumsphere $=100^{\circ} C$
Final temperature $=23^{\circ} C$
Change in temperature $(\Delta T)=\left(100^{\circ} C -23^{\circ} C \right)=77^{\circ} C$
Let specific heat capacity of aluminium be $s_{ Al }$.
The amount of heat lost by the aluminium sphere $=m_{1} s_{A l} \Delta T=0.047 kg \times s_{A l} \times 77^{\circ} C$
Mass of water $\left(m_{2}\right)=0.25 kg$
Mass of calorimeter $\left(m_{3}\right)=0.14 kg$
Initial temperature of water and calorimeter=20 $^{\circ} C$
Final temperature of the mixture $=23^{\circ} C$
Change in temperature $\left(\Delta T_{2}\right)=23^{\circ} C -20^{\circ} C =3^{\circ} C$
Specific heat capacity of water $\left(s_{w}\right)$
$=4.18 \times 10^{3} J kg ^{-1} K ^{-1}$
Specific heat capacity of copper calorimeter $=0.386 \times 10^{3} J kg ^{-1} K ^{-1}$
The amount of heat gained by water and calorimeter $=m_{2} s_{w} \Delta T_{2}+m_{3} s_{ ca } \Delta T_{2}$
$=\left(m_{2} s_{w}+m_{3} s_{ cu }\right)\left(\Delta T_{2}\right)$
$=\left(0.25 kg \times 4.18 \times 10^{3} J kg ^{-1} K ^{-1}+0.14 kg \times\right.$
$\left.0.386 \times 10^{3} J kg ^{-1} K ^{-1}\right)\left(23^{\circ} C -20^{\circ} C \right)$
In the steady state heat lost by the aluminium sphere $=$ heat gained by water + heat gained by calorimeter. So, $0.047 kg \times s_{ A 1} \times 77^{\circ} C$
$=\left(0.25 kg \times 4.18 \times 10^{3} J kg ^{-1} K ^{-1}+0.14 kg \times\right.$
$\left.0.386 \times 10^{3} J kg ^{-1} K ^{-1}\right)\left(3^{\circ} C \right)$
$s_{A t}=0.911 kJ kg ^{-1} K ^{-1}$
Similar Questions
$20\, gm$ of boiling water is poured into an ice-cold brass vessel (specific heat $0.1\, cal/gm-\,^oC$) of mass $100\, gm$. The resulting temperature is ........ $^oC$
$20\, gm$ of boiling water is poured into an ice-cold brass vessel (specific heat $0.1\, cal/gm-\,^oC$) of mass $100\, gm$. The resulting temperature is ........ $^oC$
When $x\, grams$ of steam at $100\,^oC$ is mixed with $y\,grams$ of ice at $0\,^oC$ , We obtain $(x + y)\,grams$ of water at $100\,^oC$ . What is the ratio $y/x$ ?
When $x\, grams$ of steam at $100\,^oC$ is mixed with $y\,grams$ of ice at $0\,^oC$ , We obtain $(x + y)\,grams$ of water at $100\,^oC$ . What is the ratio $y/x$ ?
Ice at $-20\,^oC$ is added to $50\,g$ of water at $40\,^oC.$ When the temperature of the mixture reaches $0\,^oC,$ it is found that $20\,g$ of ice is still unmelted. The amount of ice added to the water was close to ........$g$ (Specific heat of water $= 4.2\,J/g/^oC)$ Heat of fusion of water at $0^oC = 334\,J/g$ )
Ice at $-20\,^oC$ is added to $50\,g$ of water at $40\,^oC.$ When the temperature of the mixture reaches $0\,^oC,$ it is found that $20\,g$ of ice is still unmelted. The amount of ice added to the water was close to ........$g$ (Specific heat of water $= 4.2\,J/g/^oC)$ Heat of fusion of water at $0^oC = 334\,J/g$ )
- [JEE MAIN 2019]
Two liquids $A$ and $B$ are at $32\,^oC$ and $24\,^oC.$ When mixed in equal masses the temperature of the mixture is found to be $28\,^oC$. Their specific heats are in the ratio of
Two liquids $A$ and $B$ are at $32\,^oC$ and $24\,^oC.$ When mixed in equal masses the temperature of the mixture is found to be $28\,^oC$. Their specific heats are in the ratio of
$10\,gm$ of ice at $0\,^oC$ is mixed with $'m'\,gm$ of water at $50\,^oC$ . ........ $gm$ is minimum value of $m$ so that ice melts completely. ( $L_f = 80\,cal/gm$ and $S_W = 1\,cal/gm-\,^oC$ )
$10\,gm$ of ice at $0\,^oC$ is mixed with $'m'\,gm$ of water at $50\,^oC$ . ........ $gm$ is minimum value of $m$ so that ice melts completely. ( $L_f = 80\,cal/gm$ and $S_W = 1\,cal/gm-\,^oC$ )