Question

17 grams of ice at –36°C is to be changed to steam at 178°C. The
entire process requires _____ cal. *Round your answer to the
nearest whole number. *The specific heat of both ice
and steam is 0.5 cal/g°C. The specific heat of water is 1.00
cal/gK. The heat of fusion is 80 cal/g and the heat of vaporization
is 540 cal/g.

Answer #1

What mass of steam at 160 degree C must be mixed with 150g of ice
at 0 degree C, in a thermally insulated container, to produce
liquid water at 60 degrees C? Specific heat (heat capacity) of
water is 1 cal/gC, of steam 0.48 cal/gC and of ice 0.5 cal/gC.
Latent heat of melting for ice is 80 cal/g. Latent heat of
vaporization for steam is 540 cal/g.

The heat of fusion of water is 79.9 cal/g, the heat of
vaporization of water is 540 cal/g, and the specific heat of water
is 1.00 cal/deg/g. How many grams of ice at 0 ° could be converted
to steam at 100 °C by 9,076 cal of heat?

10.0g of ice at -0.50c is placed into a 100g aluminum cup at
300C. the latent heat of vaporization of water is 540 cal/g. the
latent of fusion of water is 80 cal/g. the specific heat of water
is 1.00 cal/gC. the specific heat of ice is 0.500cal/gC. the
specific heT of steam is 0.216 cal/gC. which of the following
sutuations is correct
A) T(equil)=97C with 10.0g of water
B) t(equil)=100C w/4.6g steam
C) t(equil)=152C w/10g steam
D) T(equil)=100C w/6.4g...

Calculate the heat required to convert 10.0 g of ice at 0.0 °C
to steam at 100.0 °C. The specific heat of water is 1.00 cal/(g x
°C); the heat of fusion is 80.0 cal/g; and the heat of vaporization
is 540.0 cal/g.
Choose one of the following
7.20 x 103 cal
1.80 x 103 cal
5.40 x 103 cal
6.20 x 103 cal
6.40 x 103 cal

To change 25 kg of ice -10°C to steam 100°C, how much heat is
required? The specific heat of water is 4.184 kJ/kg. K. The latent
heat of fusion for water at 0°C is approximately 334 kJ/kg (or 80
cal/g), and the latent heat of vaporization at 100°C is about 2,230
kJ/kg (533 cal/g).

Im taking a course to review general chemistry. In lecture, my
professor didnt say how to work this kind of problem in the
homework. Please list all the steps. Thanks.
The heat of fusion of water is 79.9 cal/g, the heat of
vaporization of water is 540 cal/g, and the specific heat of water
is 1.00 cal/deg/g. How many grams of ice at 0 ° could be converted
to steam at 100 °C by 9,958 cal of heat?

2. Sixty grams of water is at an initial temperature of 24
°C.
Calculate the heat required to completely convert the 100 ºC
water to steam. (The latent heat of vaporization of water is 540
cal/g.)
The heat required to completely convert the 100 ºC water to
steam is... kcal?
3. How much heat is required to raise the temperature of 140 g
of water from 12°C to 88°C? The specific heat capacity of water is
1 cal/g·°C.
The heat...

suppose the specific heat of ice and water is 0.49 cal/g. C ( C
represent degree Celsius) and 1.0 cal/g. C. the latent heat of
fusion of water is 80 cal/g. how much heat (in calories) is
required for 100 grams of ice with an initial temperature of -10 C
to a. raise the ice's temperature to the melting point? b. then
completely melt the ice to water? c. finally, raise the water's
temperature to 50 C?

What mass of steam at 100∘C must be added to 1.00 kg of ice at
0∘C to yield liquid water at 18 ∘C? The heat of fusion for water is
333 kJ/kg , the specific heat is 4186 J/kg⋅C∘J/kg⋅C∘ , the heat of
vaporization is 2260 kJ//kg .
Express your answer to two significant figures and include the
appropriate units
m=

If 10 g of steam of 100 C is introducded into a mixture of 200 g
of water and 120 g of ice find the the final temperature and
composition of the mixture?
mass ice= 120g, c=0.5
mass water=200 g , c=1
Lf= 80 cal/ g Lv= 540 cal

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