Question #f5a12

2 Answers
Mar 22, 2016

Answer:

=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#

Mar 23, 2016

Answer:

#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