Question

Question #bc6e9

Answer 1

(1). Amount of heat energy `Q` required to raise temperature through `DeltaT^@`of a sample of mass `m" kg"` is given by the expression

`Q=msDeltaT`
where `s` is specific heat capacity of the material of sample.

For two blocks one of copper and other of iron, are of same mass, and are raised through same temperature. We see that amount of heat required is directly proportional to specific heat capacity.

`Qprops`

`=>` block with higher specific heat capacity will require more heat energy. On comparison we see that
specific heat capacity of copper `s_C=385 J (kg ^@K)^-1`
specific heat capacity of iron `s_I=460 J (kg ^@K)^-1`
`:.` Iron block has the higher amount of energy after heating.

(2). Assuming that iron block in (1) above is immersed into `1 kg` water at `30^@ C`
Let the final temperature of water be `=30+T`
Heat gained by water`=ms_WDeltaT_W=1xx4200xx(30+T-30)`
`=4200TJ`
Final temperature of iron block is also same.
`:.`Heat lost by iron block`=ms_IDeltaT_I=1xx460xx(300-[30+T])=460(270-T)`
`=(124200-460T)J`

Equating heat gained to heat lost we get
`4200T=(124200-460T)`
`=>4200T+460T=124200`
`=>T=124200/4660=26.7^@C`, rounded to one decimal place.

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Alternatively

Let final temp of iron-water mixture be `=T`
Heat gained by water`=ms_WDeltaT_W=1xx4200xx(T-30)J`
Heat lost by iron block`=ms_IDeltaT_I=1xx460xx(300-T)`

Heat gained by water`=`Heat lost by iron block
`4200xx(T-30)=460xx(300-T)`
`=>4200T-126000=138000-460T`
`=>4660T=264000`
`=>T=56.7^@C`
Rise in temperature of water `=56.7-30=26.7^@C`, rounded to one decimal place.