Question

250 g of water at ${30}^{\circ }\mathrm{C}$ is present in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to $5^{\circ }\mathrm{C}$.(Take, specific latent heat of fusion of ice = $336\times {10}^3\mathrm{J}.{\mathrm{kg}}^{-1}$, specific heat capacity of copper vessel = $400\mathrm{J}.{\mathrm{kg}}^{-1}{}^{\circ }\mathrm{C}^{-1}$and specific heat capacity of water = $4200\mathrm{J}.{\mathrm{kg}}^{-1}{}^{\circ }\mathrm{C}^{-1}$)

Answer 1

Step 1: Given data

Mass of the copper vessel ${\mathrm{m}}_1=50\mathrm{g}$

Mass of the water ${\mathrm{m}}_2=250\mathrm{g}$

Initial temperature, $\mathrm{t}={30}^{\circ }\mathrm{C}$

Final temperature $\mathrm{T}=5^{\circ }\mathrm{C}$

Specific latent heat of fusion of ice ${\mathrm{L}}_1=336\times {10}^3\mathrm{J}.{\mathrm{kg}}^{-1}$

Specific heat capacity of copper vessel ${\mathrm{L}}_2=400\mathrm{J}.{\mathrm{kg}}^{-1}{}^{\circ }\mathrm{C}^{-1}$

Specific heat capacity of water ${\mathrm{L}}_3=4200\mathrm{J}.{\mathrm{kg}}^{-1}{}^{\circ }\mathrm{C}^{-1}$

Let the mass of the ice be $\mathrm{m}$.

Step 2: Calculation of the mass of the ice

Total heat gained = Total heat lost

$$\begin{gathered} {\mathrm{mL}}_1+{\mathrm{mL}}_3\mathrm{T}={\mathrm{m}}_2{\mathrm{L}}_3\left(\mathrm{t}-\mathrm{T}\right)+{\mathrm{m}}_1{\mathrm{L}}_2\left(\mathrm{t}-\mathrm{T}\right) \\ \mathrm{m}\times 336\times {10}^3\mathrm{J}.{\mathrm{kg}}^{-1}+\mathrm{m}\times 4200\mathrm{J}.{\mathrm{kg}}^{-1}.{}^{\circ }\mathrm{C}^{-1}\times 5{}^{\circ }\mathrm{C}=250\mathrm{g}\times 4200\mathrm{J}.{\mathrm{kg}}^{-1}.{}^{\circ }\mathrm{C}^{-1}\times 25{}^{\circ }\mathrm{C}+50\mathrm{g}\times 400\mathrm{J}.{\mathrm{kg}}^{-1}.{}^{\circ }\mathrm{C}^{-1}\times 25{}^{\circ }\mathrm{C} \\ 336\mathrm{m}\times {10}^3+21000\mathrm{m}=26.25\times {10}^6+500000 \\ 357000\mathrm{m}=26750000 \\ \mathrm{m}=\frac{26750000}{357000} \\ \mathrm{m}=74.93\mathrm{g} \end{gathered}$$

Thus the mass of ice required to bring down the temperature of the vessel and its contents to $5^{\circ }\mathrm{C}$ is 74.93 g.