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NEET Physics Thermodynamical Processes NEET PYQ-Thermodynamical Processes

  • question_answer
    A refrigerator works between \[4{{\,}^{o}}C\] and \[30{{\,}^{o}}C\]. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is: (Take 1 cal = 4.2 Joules)                                                                                                                                                      [NEET - 2016]

    A)  2.365 W          

    B)       23.65 W

    C)  236.5 W          

    D)       2365 W

    Correct Answer: C

    Solution :

    [c] \[\beta =\frac{{{Q}_{2}}}{W}=\frac{{{T}_{2}}}{{{T}_{1}}-{{T}_{2}}}\]
    (Where \[{{Q}_{2}}\] is heat removed)
    \[\Rightarrow \]   \[\frac{600\times 4.2}{W}=\frac{277}{303-277}\]
    \[\Rightarrow \]   \[W=236.5\,\text{joule}\]
    \[\Rightarrow \]  \[\text{Power}\,\,=\,\,\frac{W}{t}=\frac{236.5\,\text{joule}}{1\,\sec }=236.5\,\,watt.\]


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