Question

Question #f5a12

Answer 1

=31231J#

Explanation:

Heat required to raise the temp of ice from `-19 to 0^oC`
`H_1= 10.01xx2.06xx(0-(-19))~~391.79J` (Taking Spheat of ice`2.06J/(g^oC)`)
Heat equired to melt ice ice`H_2=10.01xx336J=3363.36J`
(Taking latent heat of fusion of ice`336J/(g)`)
Heat required to raise the temp of water from `-0^oC to 100^oC`
`H_3=10.01xx4.2xx100=4204.20J`

Heat required to vaporize water at its BP
`H_4=10.01xx2268=22702.68J`
(Taking latent heat of vaporization of water`336J/(g)`)

Heat required to raise the temp of vapor from `100^oC to 128^oC`
`H_5=10.01xx2.03xx28=568.97J`
(Taking Spheat of vapor `2.03J/(g^oC)`

TOTAL `H=31231J`

Answer 2

`7442.44cal` rounded to second place of decimal

Explanation:

We know that in such thermal interactions heat gained/lost is given by
`DeltaQ=mst`, or `DeltaQ=mL`
where `m,s and t` are the mass, specific heat and rise or gain in temperature of the object;
`L` is the latent heat for the change of state.
Lets choose CGS units as preferred system.

This a five step process.
1. Heating of ice at `-19^@"C"` to make it ice at `0^@"C"`. Taking the specific heat of ice as `0.5 cal//g ^@"C"` and `t=19`

`DeltaQ_1=10.01xx0.5xx19=95.095cal`
2. State of change from ice at `0^@"C"` to water at `0^@"C"`. Taking latent heat of fusion of ice as `80 cal //g`

`DeltaQ_2=10.01xx80=800.8cal`
3. Heating of water at `0^@"C"` to water at `100^@"C"`. Taking the specific heat of water as `1 cal//g ^@"C"` and `t=100`

`DeltaQ_3=10.01xx1xx100=1001cal`
4. State of change from water at `100^@"C"` to steam at `100^@"C"`. Taking latent heat of evaporation of water as `540 cal //g`

`DeltaQ_4=10.01xx540=800.8cal=5405.4cal`
5. Heating of steam at `100^@"C"` to steam at `128^@"C"`. Taking the specific heat of steam as `0.5 cal//g ^@"C"` and `t=28`

`DeltaQ_5=10.01xx0.5xx28=140.14cal`

Total heat required is
`DeltaQ_1+DeltaQ_2+DeltaQ_3+DeltaQ_4+DeltaQ_5=7442.435cal`

`=7442.44cal` rounded to second place of decimal