Question

A copper cube measuring 1.58 cm on edge and an aluminum cube measuring 1.68 cm on edge are both heated to 58.7 ∘C and submerged in 100.0 mL of water at 21.9 ∘C. What is the final temperature of the water when equilibrium is reached? (Assume a density of 0.998 g/mL for water.)

Answer #1

A copper cube measuring 1.58 cm on edge and an aluminum cube measuring 1.68 cm on edge are both heated to 56.4 C and submerged in 100.0 mL of water at 21.6 C.

First, we need to calculate the mass of the copper cube and the mass of aluminum cube. Both can be found by using the formula for density:

density = mass / volume

or

mass = (density)(volume)

mass Cu = (8.92 g/cm^3)(1.58 cm)^3 = (8.92 g/cm^3)(3.944312 cm^3)

mass Cu = 35.183263 g

mass Al = (2.70 g/cm^3)(1.68 cm)^3 = (2.70 g/cm^3)(4.741632 cm^3)

mass Al = 12.802406 g

heat lost by both metals = heat gained by water = q

heat lost by Cu = – [(mass Cu)(specific heat of Cu)(final temperature Cu - initial temperature Cu)

heat lost by Al = [(mass Al)(specific heat Al)(final temperature Al - initial temperature Al)

heat gained by H2O = [(mass H2O)(specific heat H2O)(final temperature H2O - initial temperature H2O)]

Note that the final temperature is the same for all.

– q of metals = + q of water

– [(mass Cu)(specific heat Cu)(final temperature Cu – initial temperature Cu) + (mass Al)(specific heat Al)(final temperature Al – initial temperature Al) = [(mass H2O)(specific heat H2O)(final temperature H2O – initial temperature H2O)]

– [(35.183263 g)(0.386 j/g•°C)(final temperature – 56.4°C) + (12.802406 g)(0.900 j/g•°C)(final temperature – 56.4°C) = [(100.0 g)(4.186 j/g•°C)(final temperature - 21.6 °C)]

– [(13.580735 J/°C)(final temperature – 56.4°C) + (11.522165 J/°C)(final temperature – 56.4°C) = [(418.6 J/°C)(final temperature – 21.6°C)]

– [13.580735 J/°C (final temperature) – 765.95345 J + 11.522165 (final temperature) – 649.850 J] = [418.6 J/°C (final temperature) – 9041.76 J]

– 13.580735 J/°C (final temperature) + 765.95345 J – 11.522165 (final temperature) + 649.850 J = 418.6 J/°C (final temperature) – 9041.76 J

9041.76 J + 765.95345 J + 649.850 J = 13.580735 J/°C (final temperature) + 11.522165 J/°C (final temperature) + 418.6 J/°C (final temperature)

10457.663 J = 443.7025 J/°C (final temperature)

final temperature = (10457.663 J) / (443.7025 J/°C)

final temperature = 23.6367°C or 23.6°C rounded to three significant figures

A copper cube measuring 1.60 cm on edge and an aluminum cube
measuring 1.64 cm on edge are both heated to 55.3 ∘C and submerged
in 100.0 mL of water at 21.5 ∘C.What is the final temperature of
the water when equilibrium is reached? (Assume a density of 0.998
g/mL for water.)

A copper cube measuring 1.60 cm on edge and an aluminum cube
measuring 1.64 cm on edge are both heated to 55.3 ∘C and submerged
in 100.0 mL of water at 21.5 ∘C.
What is the final temperature of the water when equilibrium is
reached? (Assume a density of 0.998 g/mL for water.)

A pure gold ring and pure silver ring have a total mass of 15.1
g . The two rings are heated to 67.4 ∘C and dropped into a 12.7 mL
of water at 22.8 ∘C. When equilibrium is reached, the temperature
of the water is 24.9 ∘C.
Part A: What is the mass of gold ring? (Assume a density of
0.998 g/mL for water.)
Part B: What is the mass of silver ring? (Assume a density of
0.998 g/mL for...

6. A cube of aluminum 10.0 cm on each side is cooled from 111 °C
to 25 °C. If the energy removed from the aluminum cube were added
to a copper cube of the same size at 35 °C, what would be the final
temperature of the copper cube?(ρAl = 2700 kg/m³; ρcopper = 8890
kg/m³)

A pure gold ring and pure silver ring have a total mass of 16.2
g . The two rings are heated to 78.9 ∘C and dropped into a 15.0 mL
of water at 21.3 ∘C. When equilibrium is reached, the temperature
of the water is 23.8 ∘C.
What is the mass of gold ring? (Assume a density of 0.998 g/mL
for water.)
What is the mass of silver ring? (Assume a density of 0.998 g/mL
for water.)
Gosh, I cant...

A small cube of lithium (density = 0.535 g/cm3) measuring 1.0 mm
on each edge is added to 0.550 L of water.The following reaction
occurs:
2Li(s)+2H2O(l)→2LiOH(aq)+H2(g)
What is the freezing point of the resulting solution? Assume
full dissociation of strong electrolytes and negligible solubility
of hydrogen gas in water. The molal freezing-point-depression
constant for water is 1.86°C/m

A small cube of lithium (density = 0.535 g/cm3) measuring 1.0 mm
on each edge is added to 0.550 L of water.The following reaction
occurs: 2Li(s)+2H2O(l)→2LiOH(aq)+H2(g)
What is the freezing point of the resulting solution? Assume
full dissociation of strong electrolytes and negligible solubility
of hydrogen gas in water. The molal freezing-point-depression
constant for water is 1.86°C/m.
I got -0.517 C but this is incorrect

Imagine that you place a piece of cork measuring 1.30 cm x 5.50
cm x 3.0 cm in a pan of WATER and that on top of the cork you place
a cube of lead measuring 1.15 cm on each edge. The density of the
cork is 0.235 g/cm^3, and the density of the lead is 11.35 g/cm^3.
Will the combination of the cork and lead float or sink? Show your
work.

A 40.0 g Ice Cube floats in 200.0 g of water in a 100.0 g copper
cup! All are at a temperature of 0.000C. Then, a piece of lead at
98.00C is dropped into the cup, and the final equilibrium
temperature is 12.00C. The specific heat of water and ice is 1.00
cal/g 0C, the specific heat of copper is 0.0924 cal/g 0C, the
specific heat of lead is 0.0305 cal/g 0C, and the heat of fusion of
ice is...

An aluminum cup contains 225 g of water and a 40 g copper
stirrer, all at 27°C. A 470 g sample of silver at an initial
temperature of 89°C is placed in the water. The stirrer is used to
stir the mixture gently until it reaches its final equilibrium
temperature of 32°C. Calculate the mass of the aluminum cup.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 15 minutes ago

asked 27 minutes ago

asked 35 minutes ago

asked 37 minutes ago

asked 43 minutes ago

asked 51 minutes ago

asked 56 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago