question_answer

Work done in converting one gram of ice at ?10°C into steam at 100°C is [MP PET/PMT 1988; EAMCET (Med.) 1995; MP PMT 2003]

A)            3045 J                                      

B)            6056 J

C)            721 J                                        

D)            616 J

Correct Answer: A

Solution :

                   Ice (?10°C) converts into steam as follows                    (ci = Specific heat of ice, cW = Specific heat of water)                                       Total heat required \[Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}\]                    Þ \[Q=1\times 0.5(10)+1\times 80+1\times 1\times (100-0)+1\times 540\]                            \[=725\,cal\]            Hence work done \[W=JQ=4.2\times 725=3045\,J\]