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sergeinik [125]
3 years ago
7

Please help!

Chemistry
1 answer:
lyudmila [28]3 years ago
4 0
<h3>The equation</h3>

4.94 x c . (Tm-23.6)=C.(23-6-21.9)+123.13 x c. (23.6-21.9)

<h3>Further explanation </h3>

The law of conservation of energy can be applied to heat changes, ie the heat received/absorbed is the same as the heat released  

Q in = Q out  

Heat can be calculated using the formula:  

Q = mc∆T  

A calorimeter is a device used to measure the specific heat of material  

A metal is put into a calorimeter that contains water and there will be heat transfer:  

\tt \displaystyle m_mc_m (T_m-T)=m_wc_w(T-Tw)

m = metal  

w = water  

T = the final temperature of the mixture  

mass of metal =(Nickel) = 4.94 g

mass of calorimeter = 12.5 g

mass of water = 123.13 g (135.63 - 12.5)

The equation

Q released (metal) = Q absorbed(calorimeter+water)

  • Qmetal = 4.94 x c . (Tm-23.60)
  • Q calorimeter = C.(23-6-21.9) --> C = heat capacity of calorimeter
  • Q water = 123.13 x c. (23.6-21.9)

The equation :

4.94 x c . (Tm-23.6)=C.(23-6-21.9)+123.13 x c. (23.6-21.9)

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<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

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  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

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