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patriot [66]
3 years ago
12

10.0 grams of water are heated during the preparation of a cup of coffee 1.0x 103 j of the heat are added to the water. which is

initially at 20 c what is the final temperature of the coffee
Chemistry
1 answer:
katovenus [111]3 years ago
6 0

<u>Answer:</u> The final temperature of the coffee is 43.9°C

<u>Explanation:</u>

To calculate the final temperature, we use the equation:

q=mC(T_2-T_1)

where,

q = heat released = 1.0\times 10^3J=1000J

m = mass of water = 10.0 grams

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

T_2 = final temperature = ?

T_1 = initial temperature = 20°C

Putting values in above equation, we get:

1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC

Hence, the final temperature of the coffee is 43.9°C

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Potassium nitrate has a lattice energy of -163.8 kcal/mol and a heat of hydration of -155.5 kcal/mol.
Brut [27]

Answer:

<em>293.99 g </em>

OR

<em>0.293 Kg</em>

Explanation:

Given data:

Lattice energy of Potassium nitrate (KNO3) = -163.8 kcal/mol

Heat of hydration of KNO3 = -155.5 kcal/mol

Heat to absorb by KNO3 = 101kJ

To find:

Mass of KNO3 to dissolve in water = ?

Solution:

Heat of solution = Hydration energy - Lattice energy

                           = -155.5 -(-163.8)

                           = 8.3 kcal/mol

We already know,

1 kcal/mol = 4.184 kJ/mole

Therefore,

= 4.184 kJ/mol x 8.3 kcal/mol

= 34.73 kJ/mol

Now, 34.73 kJ of heat is absorbed when 1 mole of KNO3 is dissolved in water.

For 101 kJ of heat would be

= 101/34.73

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Molar mass of KNO3 = 101.1 g/mole

Mass of KNO3 = Molar mass x moles

                         = 101.1 g/mole  x  2.908

                         = 293.99 g

                         = 0.293 kg

<em><u>293.99 g potassium nitrate has to dissolve in water to absorb 101 kJ of heat. </u></em>     

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if 100. mL of 0.800 M Na2SO4 is added to 200. mL of 1.20 M NaCl, what is the concentration of Na+ ions in the final solution? As
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