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anastassius [24]
3 years ago
10

when 22 grams of water undergoes a temeprature change from 35 ˚C to 65 ˚C, how much energy does the water absorb? The specific h

eat of water is 4.184.
Chemistry
1 answer:
aalyn [17]3 years ago
7 0

Answer: 2761.44 joules

Explanation:

The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of oil = 22g

C = 4.184Jg°C

Φ = (Final temperature - Initial temperature)

= 65°C - 35°C = 30°C

Then, Q = MCΦ

Q = 22g x 4.184Jg°C x 30°C

Q = 2761.41 Joules

Thus, water absorb 2761.44 joules of heat.

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At exactly 2:00 PM, speedy the hermit crab crawl onto a meter stick starting at the 10 cm mark. If he reaches the 60 cm mark at
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Answer:

speed=5cm/minute

Explanation:

Given at 2:00PM the hermit is at 10cm mark

Also given that the hermit reaches 60cm mark at 2:10PM

Therefore the time elapsed is  10 minutes = 10\times 60=600 seconds

the distance travelled by the hermit is 60-10=50cm=0.5m

We know that

speed=\frac{distane}{time}

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3 years ago
1. The molar mass of BeF2
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A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree
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Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = T_1=33.9^oC=33.9+273K=306.9 K

Final temperature of the water = T_2

Specific heat capacity of water under these conditions =  c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)

-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC

The new temperature of the water bath 32.0°C.

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The amount of space an object occupies.

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