A) \[1{}^\circ C\]
B) \[4.2{}^\circ C\]
C) \[10{}^\circ C\]
D) \[42{}^\circ C\]
Correct Answer: C
Solution :
\[Q=mc\Delta \theta \] or \[420=10\times {{10}^{-3}}\times (4.2\times 1000)\Delta \theta \] (Specific heat of water \[=4.2\times kJ\,K{{g}^{-1}}\,{{C}^{-1}}\]) \[\Delta \theta =\frac{420}{10\times {{10}^{-3}}\times 4200}={{10}^{o}}C\]
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