Question

Calculate the energy needed to heat **14.6** g ice
at **-10.0** °C to liquid water at
**70.0** °C. The heat of vaporization of water = 2257
J/g, the heat of fusion of water = 334 J/g, the specific heat
capacity of water = 4.18 J/g·°C, and the specific heat capacity of
ice = 2.06 J/g·°C.

Answer #1

**Ti = -10.0**

**Tf = 70.0**

**here**

**Cs = 2.06 J/g.oC**

**Heat required to convert solid from -10.0 to
0.0**

**Q1 = m*Cs*(Tf-Ti)**

**= 14.6 g * 2.06 J/g.oC *(0-(-10)) oC**

**= 300.76 J**

**Lf = 334.0**

**Heat required to convert solid to liquid at
0.0**

**Q2 = m*Lf**

**= 14.6 g *334.0 J/g**

**= 4876.4 J**

**Cl = 4.18 J/g.oC**

**Heat required to convert liquid from 0.0 to
70.0**

**Q3 = m*Cl*(Tf-Ti)**

**= 14.6 g * 4.18 J/g.oC *(70-0) oC**

**= 4271.96 J**

**Total heat required = Q1 + Q2 + Q3**

**= 300.76J + 4876.4J + 4271.96 J**

**= 9449 J**

**= 9.45*10^3 J (Taking into account significant
figure)**

**Answer: 9.45*10^3 J**

The enthaply of fusion of ice is 334 J/g. The heat capacity of
liquid water is 4.18 j/gxC. What is the smallest number of ice
cubes at 0C, each containing one mole of water necessary to cool
500 g of liquid water intially at 20 C to 0 C?

Calculate the amount of energy (in kJ) required to heat
10.0 g of water from 50.0°C to 150.°C at constant pressure.
(specific heat capacity of liquid water is 4.18
J/g⋅K; specific heat capacity of water vapor is
1.84 J/g⋅K; heat of vaporization of water is 2.260
kJ/g).
(1) 16.2 kJ (2) 25.6 kJ (3) 5.4 kJ (4) 33.2 kJ (5) 1.6
kJ
I know that the answer is (2) 25.6 KJ but I do not know how to
get to...

The constants for H2O are shown here:
Specific heat of ice: sice=2.09 J/(g⋅∘C)
Specific heat of liquid water: swater=4.18 J/(g⋅∘C)
Enthalpy of fusion (H2O(s)→H2O(l)): ΔHfus=334 J/g
Enthalpy of vaporization (H2O(l)→H2O(g)): ΔHvap=2250 J/g
How much heat energy, in kilojoules, is required to convert 36.0
g of ice at −18.0 ∘C to water at 25.0 ∘C ?

How much heat energy, in kilojoules, is required to convert 46.0
g of ice at −18.0 ∘C to water at 25.0 ∘C ? Express your answer to
three significant figures and include the appropriate units.
The constants for H2O are shown here:
Specific heat of ice: sice=2.09 J/(g⋅∘C)
Specific heat of liquid water: swater=4.18 J/(g⋅∘C)
Enthalpy of fusion (H2O(s)→H2O(l)): ΔHfus=334 J/g
Enthalpy of vaporization (H2O(l)→H2O(g)): ΔHvap=2250
J/g

How much heat energy, in kilojoules, is required to convert 79.0
g of ice at −18.0 ∘C to water at 25.0 ∘C ? Express your answer to
three significant figures and include the appropriate units.
The constants for H2O are shown here:
Specific heat of ice: sice=2.09 J/(g⋅∘C)
Specific heat of liquid water: swater=4.18 J/(g⋅∘C)
Enthalpy of fusion (H2O(s)→H2O(l)): ΔHfus=334 J/g
Enthalpy of vaporization (H2O(l)→H2O(g)): ΔHvap=2250
J/g

part A How much heat energy, in
kilojoules, is required to convert 69.0 g of ice at −18.0 ∘C to
water at 25.0 ∘C ? Part B How long would it take
for 1.50 mol of water at 100.0 ∘C to be converted completely into
steam if heat were added at a constant rate of 22.0 J/s ?
Specific heat of ice: sice=2.09 J/(g⋅∘C)
Specific heat of liquid water: swater=4.18 J/(g⋅∘C)
Enthalpy of fusion (H2O(s)→H2O(l)): ΔHfus=334 J/g
Enthalpy of vaporization (H2O(l)→H2O(g)):...

How much energy (in kilojoules) is needed to heat 4.65 g of ice
from -11.5 ∘C to 20.5 ∘C? The heat of fusion of water is
6.01kJ/mol, and the molar heat capacity is 36.6 J/(K⋅mol) for ice
and 75.3 J/(K⋅mol) for liquid water.

100. g of ice at 0 degrees C is added to 300.0 g of water at 60
degrees C. Assuming no transfer of heat to the surroundings, what
is the temperature of the liquid water after all the ice has melted
and equilibrium is reached?
Specific Heat (ice)= 2.10 J/g C
Specific Heat (water)= 4.18 J/g C
Heat of fusion = 333 J/g
Heat of vaporization= 2258 J/g

What amount of thermal energy (in kJ) is required to convert 220
g of ice at -18 °C completely to water vapour at 248 °C? The
melting point of water is 0 °C and its normal boiling point is 100
°C. The heat of fusion of water is 6.02 kJ mol-1 The heat of
vaporization of water at its normal boiling point is 40.7 kJ mol-1
The specific heat capacity of ice is 2.09 J g-1 °C-1 The specific
heat...

Calculate the amount of heat required to change 35.0 g ice at
-25.0ºC to steam at 125ºC. (Heat of fusion = 333 J/g; heat of
vaporization = 2260 J/g; specific heats: ice = 2.09 J/g·K, water =
4.18 J/g·K, steam = 1.84 J/g·K)

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 3 minutes ago

asked 18 minutes ago

asked 20 minutes ago

asked 22 minutes ago

asked 22 minutes ago

asked 28 minutes ago

asked 28 minutes ago

asked 33 minutes ago

asked 37 minutes ago

asked 38 minutes ago

asked 38 minutes ago

asked 40 minutes ago