Question

A 15.3 gram piece of lead at 115.1 °C is placed in a calorimeter containing water at 21.9 °C. If the temperature at equilibrium is 24.5 °C, what is the mass of the water?

Report your answer in grams to **two (2) sig
figs**.

Answer #1

Heat relesed by Lead ,q_{r} = m × ∆T × C

mass of lead,m = 15.3g

Temperature difference of lead , ∆T = 24.5℃ - 115.1℃ = - 90.6℃

Heat capacity of Lead ,C = 0.128J/g℃

q_{r = 15.3g ×( -
90.6℃ ) × 0.128 ( J/g ℃ )}

_{
= - 177.4J}

_{Heat released
by lead = Heat absorbed by water}

_{ Heat absorbed by water (
q}_{a) =} m × ∆T × C

mass of water ,m = ?

Temperature difference of water , ∆T = 24.5℃ - 21.9℃ = 2.6℃

Heat capacity of water, C = 4.184( J/g ℃)

177.4 J = m × 2.6℃ × 4.184( J/g ℃ )

m = 177.4J/2.6℃ × 4.184(J/g℃)

= 16.31g

so, mass of water = 16g

A 39.9 gram piece of lead at 98.7 °C is placed in a calorimeter
containing water at 21.7 °C. If the temperature at equilibrium is
27.6°C, what is the mass of the water?

A 10.67-gram -23.0°C piece of ice drops into a calorimeter
containing 150.00-gram 93.7 °C of the water. What is the final
temperature? (Swater = 4.184 J/g° C, Sice = 2.082 J/g° C, Heat of
fusion= 334.7 J/g)

A 50 gram piece of copper at 200°C is placed in 100 grams of
water at 25°C. Assuming no loss of heat to the surroundings,
determine the final temperature of the water and copper.

A calorimeter contains 82.4 grams of water at 20.9 °C. A 156
-gram piece of an unknown metal is heated to 81.9 °C and dropped
into the water. The entire system eventually reaches 26.6 °C.
Assuming all of the energy gained by the water comes from the
cooling of the metal—no energy loss to the calorimeter or the
surroundings—calculate the specific heat of the metal. The specific
heat of water is 4.18 J/g · °C _____J/g · °C

A lead (Pb) pellet having a mass of 26.47 g at 89.98°C was
placed in a constant-pressure calorimeter of negligible heat
capacity containing 100.0 mL of water. The water temperature rose
from 22.50°C to 23.17°C. What is the specific heat of the lead
pellet?

There is 300 grams of water in the 200-gram calorimeter cup
(inner can). Both are at room temperature of 20 degrees Celsius. A
100-gram metal sample with an initial temperature of 90 degrees
Celsius is placed in the calorimeter resulting. The resulting final
temperature of the system is 30 degrees Celsius. If the calorimeter
can has a specific heat of 0.2 cal/g-C degree, determine the
specific heat of the metal sample

When a 290-g piece of iron at 170 ∘C is placed in a 95-gg
aluminum calorimeter cup containing 250 g of liquid at 10∘C, the
final temperature is observed to be 32 ∘C. The value of specific
heat for iron is 450 J/kg⋅C∘, and for aluminum is 900 J/kg⋅C∘
Part A
Determine the specific heat of the liquid.
Express your answer using two significant figures.

When a 290-g piece of iron at 180 ∘C is placed in a 95-g
aluminum calorimeter cup containing 250 g of liquid at 10∘C, the
final temperature is observed to be 34 ∘C. The value of specific
heat for iron is 450 J/kg⋅C∘, and for aluminum is 900 J/kg⋅C∘.
Determine the specific heat of the liquid.
Express your answer using two significant figures.

A 102 g piece of ice at 0.0°C is placed in an insulated
calorimeter of negligible heat capacity containing 100 g of water
at 100°C. Find the entropy change of the universe for this
process?
135 J/K 134 J/K is wrong,

A 35.7 gram sample of iron (heat capacity 0.45 g/J°C) was heated
to 99.10 °C and placed into a coffee cup calorimeter containing
42.92 grams of water initially at 15.15 °C. What will the final
temperature of the system be? (Specific heat of water is 4.184
J/g°C).
Please show work.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 1 minute ago

asked 21 minutes ago

asked 21 minutes ago

asked 27 minutes ago

asked 27 minutes ago

asked 27 minutes ago

asked 29 minutes ago

asked 35 minutes ago

asked 45 minutes ago

asked 46 minutes ago

asked 46 minutes ago

asked 48 minutes ago