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alexandr402 [8]
3 years ago
13

23 grams of sodium reacts with 293 cm 3 of water that is initially at 298 K. It produces an enthalpy change of 197 kJ. What is t

he final
temperature of the water? The specific heat capacity of water is 4.18 J/Kg.
hydrogen
2
He
Chemistry
2 answers:
muminat3 years ago
5 0

Answer:

THE FINAL TEMPERATURE OF WATER IS 459.32 K

Explanation:

Mass of sodium = 23 g

Volume of water = 293 cm3

Initial temperature = 298 K

Specific heat capacity of water = 4.18 J/Kg

Enthalpy change = 197 kJ = 197, 000 J

Heat  = mass * specific heat * change in temperature

From the statement it shows that the initial temperature is 298 K or 25 °C and at this temperature the density of water is 0.99707 g/mL

Hence we can calculate the mass of water by using this formula:

Mass = volume * density

Mass = 293 cm3 * 0.99707 g/mL

Mass = 292.14 g

The enthalpy change = mass * specific heat * change in temperature

197, 000 = 292.14 * 4.18 * change in temperature

Change in temperature = 197 000 / 292.14 * 4.18

Change in temperature = 197 000 / 1221.1452

Change in temperature = 161.32 K

Hence, the final temperature of the water is therefore the addition of initial temperature and the change in temperature

Final temperature = initial temp + change in temp

Final temp = 298 K + 161.32 K

Final temp = 459.32 K

In conclusion, the final temperature of water is 459.32 K

Scorpion4ik [409]3 years ago
5 0

Answer:

458.85 K

Explanation:

Founder's Education/Educere

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