A) \[65{}^\circ C\]
B) \[70{}^\circ C\]
C) \[60{}^\circ C\]
D) \[75{}^\circ C\]
Correct Answer: A
Solution :
Let the final temperature of mixture be t. Heat lost by water at \[80{}^\circ C\] \[=m\,s\,\Delta \,t\] \[=0.1\times {{10}^{3}}\times {{s}_{water}}\times (80{}^\circ -t)\] Here \[m=v\times d=0.1\times {{10}^{3}}\,kg\] Heat gained by water at\[60{}^\circ C\] \[=0.3\times {{10}^{3}}\times {{s}_{water}}\times (t-60{}^\circ )\] By the principle of calorimetry Heat lost = Heat gained \[\therefore \]\[0.1\times {{10}^{3}}\times {{s}_{water}}\times (80-t)=\] \[0.3\times {{10}^{3}}\times {{s}_{water}}\times (t-60{}^\circ )\] or \[\left( 80{}^\circ -r \right)=3\times \left( t-60{}^\circ \right)\] or \[4t=260{}^\circ \] \[\Rightarrow \] \[t=65{}^\circ C.\]
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