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Heat capacity of water is 18 cal-"degree...

Heat capacity of water is `18 cal-"degree"^(-1)-mol^(-1)`. The quantity of heat needed to rise temperature of 18 g water by `0.2^(@)C` is x cal. Then amount of `CH_(4(g))` to be burnt to produce X cal heat is
`(CH_(4)+2O_(2) rarr CO_(2)+2H_(2)O, Delta H = -200 K.Cal)`

A

`1.8xx10^(-3)` mol

B

`3.6xx10^(-5)` mol

C

`0.0288g`

D

`0.288mg`

Text Solution

Verified by Experts

The correct Answer is:
D
`DeltaU=Zxthetax(M)/(W),DeltaH=DeltaU+DeltanRT,Q=mst`
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