A) \[1.50\,\,\times \,\,{{10}^{3}}\]
B) \[2.55\,\,\times \,\,{{10}^{3}}\]
C) \[3.00\,\,\times \,\,{{10}^{3}}\]
D) \[5.10\,\,\times \,\,{{10}^{3}}\]
Correct Answer: B
Solution :
\[{{m}_{1}}{{s}_{1}}\Delta t+{{m}_{2}}{{s}_{2}}\Delta t=\text{Work}\,\text{done}\] \[{{m}_{1}}{{s}_{1}}\Delta t+{{m}_{2}}{{s}_{2}}\Delta t={{P}_{1}}{{t}_{1}}\] where \[{{m}_{1}}=0.5\,kg\] Specific heat \[{{s}_{1}}=4200\,J/kg-K\] \[\Delta t=\Delta {{t}_{1}}=\Delta {{t}_{2}}=3K\] \[{{P}_{1}}={{P}_{2}}=10\,W\] \[{{t}_{1}}=15\times 60=900\,s\] \[{{s}_{2}}=\]Specific heat capacity of container \[0.5\times 4200\times (3-0)+{{m}_{2}}{{s}_{2}}\times (3-0)\] \[=10\times 15\times 60\] \[2100\times 3+{{m}_{2}}{{s}_{2}}\times 3=9000\] \[{{m}_{2}}{{s}_{2}}=\frac{9000-6300}{3}\] \[{{m}_{2}}{{s}_{2}}=900\] Similarly, in case of oil \[{{m}_{1}}{{s}_{0}}\Delta t+{{m}_{2}}{{s}_{2}}\Delta t={{P}_{2}}{{t}_{2}}\] where \[{{s}_{0}}=\]specific heat capacity of oil \[{{P}_{1}}={{P}_{2}}=10\,W\] \[2\times {{s}_{0}}\times 2+900\times 2=10\times 20\times 60\] \[4{{s}_{0}}+1800=12000\] \[4{{s}_{0}}=12000-1800\] \[{{s}_{0}}=\frac{10200}{4}=2550\] \[=2.55\times {{10}^{3}}\,J\,k{{g}^{-1}}{{K}^{-1}}\]
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