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krok68 [10]
3 years ago
6

How many joules of heat are given off when 5.00g of water cool from 348.0K to 298.0K?

Chemistry
1 answer:
forsale [732]3 years ago
6 0

Answer:

Q =  -1045 J

Explanation:

Given data:

Mass of water = 5.00 g

Initial temperature = 348.0 K

Final temperature = 298.0 K

Heat given off = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 J/g.K

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 = 298.0 K - 348.0 K

ΔT = - 50 K

Q = 5.0 g ×4.18 J/g.K× - 50 K

Q =  -1045 J

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Answer:The answer is D.1,3

Explanation:

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3 years ago
How many grams are in 2.45E24 formula units of Fp3BZ2? The molar mass of Fp3Bz2. is 97.05<br> g/mol.
dlinn [17]

Answer:

394.99g

Explanation:

The number of moles of a substance can be calculated by dividing the number of atoms of such substance by Avagadro's number (6.02 × 10^23)

n = nA ÷ 6.02 × 10^23

The number of atoms of Fp3BZ2 in this question is 2.45E24 formula units i.e. 2.45 × 10^24

n = 2.45 × 10^24 ÷ 6.02 × 10^23

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n = 0.407 × 10¹

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4 0
3 years ago
If 1.16 L of water is initially at 24.2 ∘C, what will its temperature be after absorption of 9.4×10−2 kWh of heat?
vitfil [10]

Answer:

The temperature will be 93.92 °C

Explanation:

To explain this we will use following equation: also  Q = ∆U + W known as the NON-FLOW ENERGY EQUATION (N.F.E.E.)

With Q = heat added to the system

with ∆U  = change in internal energy

⇒∆U = ( m )( Cv )( T2 - T1 )

With W = work done by the system

⇒For this situation W = 0 because there isn't work done

So we get: ∆U = ( m )( Cv )( T2 - T1 ) = Q

To find the temperature, we have to isolate T in the equation:

(T2-T1) = Q / (m)(Cv)

⇒ Since we know that m = density * volume we can calculate the mass of water.

mass = 1000g/L * 1.16 L = 1160g

Cv = heat capacity ⇒ water has a  heat capacity of 4.184 J/g °C

We know the absorption of heat is 9.4x 10^-2 kWh but to know how many joule this is we should convert ( 1 joule = 3.6 x 10^6 kWh)

⇒Q = ( 0.094 kWh ) ( 3.6 x 10^6 J / kWh ) = 0.3384 x 10^6 J

For the temperature we get then: T2 -T1 = Q / (m)(Cv)

T2 - T1 =  0.3384 x 10^6 J / (( 1160g)*(4.184 J/g °C)) = 69.72 ° C

T2 = ( T2 - T1 ) + T1   ⇒ 69.72 + 24.2 = 93.92 °C

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