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3 years ago

Calculate the amount of energy that is absorbed by 75 g.. water

Chemistry

1 answer:

Contact [7]3 years ago

8 0

Answer:

Q = 1147.5 cal

Explanation:

Given data:

Energy absorbed = ?

Mass of water = 75 g

Initial temperature = 22.5°C

Final temperature = 37.8°C

Specific heat of water = 1 cal/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2- T1

ΔT = 37.8°C - 22.5°C

ΔT = 15.3 °C

Q = m.c. ΔT

Q = 75 g . 1 cal/g.°C . 15.3 °C

Q = 1147.5 cal

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3 years ago

What is the uses of plastic?

Alex Ar [27]

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3 years ago

Read 2 more answers

A 48.0g sample of quartz, which has a specific heat capacity of 0.730·J·g−1°C−1, is dropped into an insulated container containi

Butoxors [25]

Answer:

The equilibrium temperature of the water is 26.7 °C

Explanation:

<u>Step 1:</u> Data given

Mass of the sample quartz = 48.0 grams

Specific heat capacity of the sample = 0.730 J/g°C

Initial temperature of the sample = 88.6°C

Mass of the water = 300.0 grams

Initial temperature = 25.0°C

Specific heat capacity of water = 4.184 J/g°C

<u>Step 2:</u> Calculate final temperature

Qlost = -Qgained

Qquartz = - Qwater

Q =m*c*ΔT

Q = m(quartz)*c(quartz)*ΔT(quartz) = -m(water) * c(water) * ΔT(water)

⇒ mass of the quartz = 48.0 grams

⇒ c(quartz) = the specific heat capacity of quartz = 0.730 J/g°C

⇒ ΔT(quartz) = The change of temperature of the sample = T2 -88.6 °C

⇒ mass of water = 300.0 grams

⇒c(water) = the specific heat capacity of water = 4.184 J/g°C

⇒ ΔT= (water) = the change in temperature of water = T2 - 25.0°C

48.0 * 0.730 * (T2-88.6) -300.0 * 4.184 *(T2 - 25.0)

35.04(T2-88.6) = -1255.2 (T2-25)

35.04T2 -3104.544 = -1255.2T2 + 31380

1290.24T2 = 34484.544

T2 = 26.7 °C

The equilibrium temperature of the water is 26.7 °C

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mr Goodwill [35]

Answer:

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Explanation:

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