Question

3.75 X105 calories of heat is given out by 5 kg of water at 100o C. Calculate

the temperature of cooled water. SHC of water =1000 calories / kg o C.

Answer 1

Given,
$$\begin{gathered} \mathrm{Heat}\mathrm{energy},\mathrm{Q}=3.75\times {10}^5\mathrm{cal} \\ \mathrm{Mass}\mathrm{of}\mathrm{water},\mathrm{m}=5\mathrm{kg} \\ \mathrm{Temperature},{\mathrm{t}}_1={100}^0\mathrm{C} \\ \mathrm{SHC}\mathrm{of}\mathrm{water},\mathrm{s}=1000\mathrm{cal}/{\mathrm{kg}}^0\mathrm{C} \\ \mathrm{Usinf}, \\ \mathrm{Q}=\mathrm{m}\times \mathrm{s}\times ∆\mathrm{t} \\ =\mathrm{m}\times \mathrm{s}\times ({\mathrm{t}}_1-{\mathrm{t}}_2) \\ 3.75\times {10}^5=5\times 1000\times (100-{\mathrm{t}}_2) \\ 100-{\mathrm{t}}_2=\frac{3.75\times {10}^5}{5000}=\frac{375}{5}=75 \\ {\mathrm{t}}_2=100-75={25}^0\mathrm{C} \\ \mathrm{Thus},\mathrm{temperature}\mathrm{of}\mathrm{cooled}\mathrm{water}\mathrm{is}{25}^0\mathrm{C}. \end{gathered}$$