Question

A quantity of ice at 0.0 °C was added to 33.6 g of water at 41.0 °C to give water at 0.0 °C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol, and the specific heat is 4.18 J/(g•°C). ______ grams

Answer #1

Heat lost by water = heat gained by ice

mcdt = m'L

Where

m = mass of water = 33.6 g

c = specific heat = 4.18 J/(g•°C)

dt = change in temperature of water = final - initial = 41.0 - 0
= 41.0 ^{o}C

m' = mass of ice = ?

L = heat of fusion of water = 6.01 kJ/mol

= 6.01x10^{3} J/mol

= 6.01x10^{3} (J/mol) x (1/18)( mol /g)

= 334 J/g

Plug the values we get

mcdt = m'L

33.6x4.18x41 = m'x334

m' = 17.2 g

Therefore the quantity of ice added is 17.2 grams

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

Calculate the amount of energy necessary to warm 10.0 g of ice
from 0.0° C to 137.0° C. Express your answer in kJ.
Do not enter units of measurement, do not enter the answer in
scientific notation.
The specific heat is 4.184 J/g∙°C for water and 1.99 J/g∙°C for
steam. ΔHvap is 40.79 kJ/mol and ΔHfus is 6.01 kJ/mol.

Calculate the enthalpy change, ΔH, for the process in
which 10.3 g of water is converted from liquid at 9.4 ∘C to vapor
at 25.0 ∘C .
For water, ΔHvap = 44.0 kJ/mol at 25.0 ∘C and
Cs = 4.18 J/(g⋅∘C) for H2O(l).
How many grams of ice at -24.5 ∘C can be completely converted to
liquid at 9.8 ∘C if the available heat for this process is
5.03×103 kJ ?
For ice, use a specific heat of 2.01 J/(g⋅∘C) and...

How much heat is released when 105 g of steam at 100.0°C is
cooled to ice at -15.0°C? The enthalpy of vaporization of water is
40.67 kJ/mol, the enthalpy of fusion for water is 6.01 kJ/mol, the
molar heat capacity of liquid water is 75.4 J/(mol • °C), and the
molar heat capacity of ice is 36.4 J/(mol • °C).

A 0.4-L glass of water at 20°C is to be cooled with ice to 5°C.
The density of water is 1 kg/L, and the specific heat of water at
room temperature is c = 4.18 kJ/kg·°C. The specific heat
of ice at about 0°C is c = 2.11 kJ/kg·°C. The melting
temperature and the heat of fusion of ice at 1 atm are 0°C and
333.7 kJ/kg.
A) Determine how much ice needs to be added to the water, in...

What is the total heat flow if 28 grams of water at 12°C is
cooled to become ice at –19°C? The specific heat of liquid water is
4.18 J/g • °C; the specific heat of ice is 2.1 J/g • °C. The heat
of fusion of ice is 333 J/g, and the freezing point of water is
0.0°C.

The enthalpy change for converting 10.0 g of ice at -25.0
degrees C to water at 80.0 degrees C is _______kJ. The
specific heats of ice, water, and steam are 2.09 J/g-K, 4.18 J/g-K,
and 1.84 J/g-K, respectively. For H2O, Delta Hfus=6.01
kJ/mol, and Delta Hvap=40.67 Kj/mol
Please explain steps used as well. Thank you.

Two 20.0-g ice cubes at –20.0 °C are placed into 285 g of water
at 25.0 °C. Assuming no energy is transferred to or from the
surroundings, calculate the final temperature, Tf, of the water
after all the ice melts. heat capacity of H2O(s) is 37.7 J/mol*K
heat capacity of H2O(l) is 75.3 J/mol*K enthalpy of fusion of H20
is 6.01 kJ/mol

Two 20.0-g ice cubes at –13.0 °C are placed into 275 g of water
at 25.0 °C. Assuming no energy is transferred to or from the
surroundings, calculate the final temperature of the water after
all the ice melts.
heat capacity of H2O(s)
37.7 J/(mol*k)
heat capacity of H2O(l)
75.3 J/(mol*k)
enthalpy of fusion of H2O
6.01 kJ/mol

Part A
Calculate the enthalpy change, ΔH, for the process in
which 44.0 g of water is converted from liquid at 7.6 ∘C to vapor
at 25.0 ∘C .
For water, ΔHvap = 44.0 kJ/mol at 25.0 ∘C and
s = 4.18 J/(g⋅∘C) for H2O(l)
Express your answer numerically in kilojoules.
Part B
How many grams of ice at -11.0 ∘C can be completely converted to
liquid at 9.4 ∘C if the available heat for this process is
5.66×103 kJ...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 26 minutes ago

asked 44 minutes ago

asked 47 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 3 hours ago

asked 3 hours ago

asked 3 hours ago

asked 3 hours ago

asked 4 hours ago