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3 years ago

if a 2.0g sample of water at 5.0celcius absorbs 21.8Joules of energy, the temprature of the sample will be raised by

1 answer:

8 0

You'd use the temperature change equation. The specific heat of water is always 4.18J/(g °C).
Equation to use: q=mCΔT

21.8=(2.0)(4.18)(Tf-5)
21.8=(8.36)(Tf-5)
21.8/8.36=Tf-5
2.61=Tf-5
2.61+5=Tf
Tf=7.61 °C

Since you want to know how many °C it raises, you wouldn't pay attention to the last 2 steps, however if you need to know the final temp, you want to go to the last step.

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Una muestra de 258.4 g de etanol (C2H5OH) se quemó en una bomba calorimétrica

Ilia_Sergeevich [38]

Answer:

Explanation:

Unclear question.

I infer you want a clear rendering, which reads;

A 258.4 g sample of ethanol (C2H5OH) was burned in a calorimetric pump using a Dewar glass. As a consequence, the water temperature rose to 4.20 ° C.

If the heat capacity of the water and the surrounding glass was 10.4 kJ / ° C, calculate the heat of combustion of one mole of ethanol.

4 0

3 years ago

what would the molarity of a solution be if you took 10 mL of a 13M stock solution and made a 300 mL solution?

nignag [31]

Answer:

0.43M

Explanation:

8 0

3 years ago

A monatomic ideal gas expands slowly to twice its original volume, doing 340 JJ of work in the process. Part APart complete Find

Dafna1 [17]

Answer:

A) if the system is isothermal then all the heat added to the system will be used to do work (since none is used to raise the temperature of the gas). The heat added will be equal to the work done = 340 J

B) change in internal energy of the system of the process is isothermal will be zero, since there is no rise in temperature.

C) an adiabatic process is one involving no heat loss or gain through the system, Therefore heat gain will be zero

D) if the process is adiabatic then there is no heat loss or gain through the system and hence there is no change in temperature. Change in internal energy will be zero

E) if the process is isobaric then, there is no work done and the total heat to the system is equal zero

F) if there is no work done, and no heat added, then the internal energy will be equal zero.

4 0

3 years ago

A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35 oC to 195 oC. Calculate the specific heat.

Alja [10]

Answer:

Question 1: 0.2J/(gºC)

Question 2: 6,000J

Question 3: 300J

Question 4: 80g

Question 5: 74ºC

Question 6: 50g

Explanation:

Question 1. A 20g piece of lead absorbs 566 joules of heat and its temperature changes from 35º oC to 195º C. Calculate the specific heat.

The thermal energy equation is:

Q = m × C × ΔT

Substitute and solve for C:

566J = 20g × C × (195ºC - 35ºC)

C = 566J / (20g × 160ºC)

C = 0.177 J/(gºC) ≈ 0.2J/(gºC)

You must round to one significant figure because one factor has one significant figure).

Qustion 2. 40g of water is heat at 40ºC and the temperature rise to 75ºC. What is the amount of heat needed for the temperature to rise? (specific heat of water is 4.184 J/gºC)

Use the thermal energy equation again:

Q = m × C × ΔT

Substitute and compute:

Q = 40g × 4.184 J/gºC × (75ºC - 40ºC)

Q = 5,857.6J

Round to one significant figure: 6,000J

Question 3. Graphite has a mass of 50g and a specific heat of 0.420 J/gºC. If graphite is cooled from 50ºC to 35ºC, how much energy was lost?

Q = m × C × ΔT

Q = 50g × 0.420J/gºC × (35ºC - 50ºC)

Q = 315J

Round to one significant figure (because 50g has one significant figure)

Q = 300J

Question 4. Iron has a specific heat of 0.712 J/gºC. A piece of iron absorbs 3000J of energy and undergoes a temperature change totaling 50ºC, What is the mass of iron?

Q = m × C × ΔT

Solve for m:

m = Q / (C × ΔT)

Substitute and compute:

m = 3,000J / (0.712J/gºC × 50ºC)

m = 84.26 g ≈ 80 g (rounded to one significant figure, because the factor 3,000J has one significant figure).

Question 5. If 400g of an unknown solution at 70ºC loses 7500 J of heat, what is the final temperature of the unknown solution. The unknown solution has a specific heat of 4.184 J/gºC.

Q = m × C × ΔT

Q is negative, since it is released.

Substitute and solve for T:

- 7,500J = 400g × 4.184J/gºC × (T - 70ºC)

T = - 7500J / 400g × 4.184J/gºC) + 70ºC

T = 74ºC

If you round to one significant figure you cannot tell the temperature difference, thus leave two significant figures.

Question 6. How many grams of water would require 9500J of heat to raise the temperature from 50ºC to 100ºC

Q = m × C × ΔT

Subsitute:

9,500J = m × 4.184J/gºC × (100ºC - 50ºC)

Solve for m and compute:

m = 9,500J / (4.184J/gºC × 50ºC)

m = 45g

Since the temperatures indicate one singificant figure, the mass should be rounded to one significant figure:

m = 50g.

8 0

2 years ago

A 25.0 mL sample of 0.123 M HCl was titrated with 18.04 mL of NaOH . What is the concentration of the NaOH ?

Blababa [14]

Answer:

0.170 M

Explanation:

M1V1=M2V2

(25.0mL)(0.123M)=(18.04mL)(x)

x= 0.170 M

* Please text me at 561-400-5105 for tutoring: I can do assignments, labs, exams, etc. :)

5 0

2 years ago