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Nana76 [90]
3 years ago
5

Calculate the specific heat capacity of a piece of ice if 1.30 kg of the wood absorbs 6.75×104 joules of heat, and its temperatu

re changes from 32 ºC to 57 ºC.
Chemistry
1 answer:
Gre4nikov [31]3 years ago
8 0

Answer:

c=2.0769\  \frac{J}{g\ \textdegree C}

Explanation:

-Specific heat capacity is given by the formula:

q=mc\bigtriangleup T

Where:

q is the heat gained or loosed by the substance

m is the mass of the substance

c is the specific heat of the substance

\bigtriangleup T is the change in temperature

#We make c the subject of the formula and substitute to solve for it:

q=mc\bigtriangleup T\\\\c=\frac{q}{m\bigtriangleup T}\\\\\bigtriangleup T=(57-32)\textdegree C=25\textdegree C\\\\\therefore c=\frac{6.75\times 10^4J}{1.3\times 1000\ g\times 25\textdegree  C}\\\\=2.0769 \ \frac{J}{g\ \textdegree C}

Hence, the specific heat capacity of the ice is 2.0769 \frac{J}{g\ \textdegree C}

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Answer:

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Explanation:

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A 7.74 L balloon is filled with water at 3.88 atm. If the balloon is squeezed into a 0.23 L beaker and does NOT burst, what is t
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Answer:

131 atm

Explanation:

To find the new pressure, you need to use Boyle's Law:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new pressure (P₂) by plugging the given values into equation and simplifying.

P₁ = 3.88 atm                       P₂ = ? atm

V₁ = 7.74 L                           V₂ = 0.23 L

P₁V₁ = P₂V₂                                                      <----- Boyle's Law

(3.88 atm)(7.74 L) = P₂(0.23 L)                       <----- Insert values

30.0312 = P₂(0.23 L)                                      <----- Simplify left side

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3 years ago
When 1.98g of a hydrocarbon is burned in a bomb calorimeter, the temperature increases by 2.06∘C. If the heat capacity of the ca
schepotkina [342]

Answer:

8.3 kJ

Explanation:

In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:

q for water:

q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g

                                      c: specific heat of water = 4.186 J/gºC

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so solving for q :

q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J

For calorimeter

q calorimeter  = C x  ΔT  where C: heat capacity of calorimeter = 69.6 ºC

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q calorimeter = 69.60J x 2.06 ºC = 143.4 J

Total heat released = 8,140 J +  143.4 J = 8,2836 J

Converting into kilojoules by dividing by 1000 we will have answered the question:

8,2836 J x 1 kJ/J = 8.3 kJ

7 0
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