. how much heat is released to change the steam at 135.2C to ice at 37.23C? mass=28.7g?

Question

badly need the answer

Answer 1

PROCESS:

Steam ---------> Water ----------> Ice
Heat is progressively released as the above phase conversions occur.

DATA:
m =28.7 g
specific heat of steam = s1 = 1.860 J/g.°C
specific heat of water = s2 = 4.186 J/g °C
specific heat of ice = s3 = 2.02 J/g °C
_________ __________ __________ __________ __________
Heat released for cooling of steam from 135.2 °C to 100 °C
= m*s1*dT
= 28.7*1.86*(135.2-100)
= 1879.0464 J
_________ ________ _________ _________ _________ ______
Latent heat of vaporization is 2260 kJ.kg-1 = 2260J /g

For conversion of 28.7 gram steam at 100 °C to water at 100 °C ,heat released = 2260*28.7 = 64862 J
______ ________ ________ ________ _______ ________ ____

Heat released for cooling of water from 100 °C to 0 °C
= m*s2*dT
= 28.7*4.186 *100
= 12013.82 J
_______ ______ _______ ________ ________ ________ _____

Latent heat of fusion of ice at 0 ºC, for example, is 334 kJ.kg^-1 = 334 J/g

For conversion of 28.7 gram water at 0 °C to ice at 0 °C ,heat released = 334*28.7 = 9585.8 J
_________ _________ __________ _________ _________ ____

Heat released for cooling of ice from 0 °C to -37.23°C
= m*s3*dT
= 28.7*2.02 *37.23
= 2158.37202 J
_______ _________ ___________ _________ __________ ____

Total heat to be released = summation of all = 90499.03842 J
=90.499 kJ

Answer 2

The water would not be ice at 37.23 C which is slightly above body temperature. Did you mean -37.23C? IF SO THEN
-Q = heat released OR
Q = -28.7g (2.01J x35.2 +40.7E3J/18 +4.18J x 100 + 6.01E3/18 +2.03 x 37.23)

Note: (-) sign indicates heat released and E = 10^ (E3 = 10^3)

Q = -90.7Kj or -90,700J...Be sure to check my math:>)

Answer 3

Enthalpies (latent heat) of fusion: 334 kJ/kg (converting water to ice)

Enthalpies (latent heat) of vaporization: 2260 kJ/kg (converting steam to water)

Heat released: mass of substance x specific heat capacity x temperature change

Quick glance at a steam table and the specific heat of saturated steam (@ 1bar) is 1.860 kJ/(kg*C)

Specific Heat of water: 4.19 kJ/(kg*C)

So, to get the temp down from 135.2C to 100C:
1.860kJ/(kg*C) * 0.0287kg * (135.2 - 100)C = 1.833 kJ

Heat of vaporization:
2260 kJ/kg * 0.028kg = 63.28kJ

From water at 100C to 37.23C:
4.19kJ/(kg*C) * 0.0287kg * (100 - 37.23)C = 7.36 kJ

Heat of fusion:
334 kJ/kg * 0.028kg = 9.352 kJ

Total heat released:
1.833 kJ + 63.28 kJ + 7.36 kJ + 9.352 kJ = 81.825 kJ