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a_sh-v [17]
3 years ago
15

1. When 400 J of heat is added to 5.6 g of olive oil at 23*C, the temperature increases to 87*C. What is the specific heat of th

e olive oil?
2. A small rock is heated and placed in a foam cup calorimeter containing 20.0 mL of water at 25.0*C. The water reaches a maximum temperature of 27.2*C. How many joules of heat are released by the rock?

3. How many kilojoules of heat are absorbed when 1.00 L of water is heated from 19*C to 82*C?
Chemistry
1 answer:
LenaWriter [7]3 years ago
5 0

Answer:

1. 1.116 J/g.°C.

2. 184.184 J.

3. 263.718 kJ.

Explanation:

<em>1. When 400 J of heat is added to 5.6 g of olive oil at 23°C, the temperature increases to 87°C. What is the specific heat of the olive oil?</em>

  • To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by olive oil (Q = 400.0 J).

m is the mass of olive oil (m = 5.60 g).

c is the specific heat capacity of lead (c = ??? J/g.°C).

ΔT is the temperature difference (final T - initial T) (ΔT =  87 °C - 23 °C = 64.0 °C).

∵ Q = m.c.ΔT

<em>∴ c = Q/m.ΔT </em>= (400.0 J)/(5.6 g)(64.0 °C) = <em>1.116 J/g.°C.</em>

<em />

<em>2. A small rock is heated and placed in a foam cup calorimeter containing 20.0 mL of water at 25.0°C. The water reaches a maximum temperature of 27.2°C. How many joules of heat are released by the rock?  </em>

  • Also, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by water (Q = ??? J).

m is the mass of water (m = d.V = (1.0 g/mL)(20.0 mL) = 20.0 g).

c is the specific heat capacity of lead (c = 4.186 J/g.°C).

ΔT is the temperature difference (final T - initial T) (ΔT =  27.2 °C - 25.0 °C = 2.2 °C).

<em>∴</em> Q = m.c.ΔT = (20.0 g)(4.186 J/g.°C)(2.2 °C) = 184.184 J.

<em></em>

<em>3. How many kilojoules of heat are absorbed when 1.00 L of water is heated from 19°C to 82°C?</em>

<em></em>

  • Also, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by water (Q = ??? J).

m is the mass of water (m = d.V = (1.0 g/mL)(1000.0 mL) = 1000.0 g = 1.0 kg).

c is the specific heat capacity of lead (c = 4.186 J/g.°C).

ΔT is the temperature difference (final T - initial T) (ΔT =  82.0 °C - 19.0 °C = 63.0 °C).

<em>∴</em> Q = m.c.ΔT = (1.0 kg)(4.186 J/g.°C)(63.0 °C) = 263.718 kJ.

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4 0
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Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

3 0
3 years ago
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