How many milliliters of 2.5M HCI are required to exactly neutralize 15 milliliters of 5.0 M NaOH
1 answer:
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42.34 g of water could be warmed from 21.4°C to 43.4°C by the pellet dropped inside it
Heat loss by the pellet is equal to the Heat gained by the water.
….(1)
where, is the heat gained by water
is the heat loss by pellet
= mCΔT
where m = mass of water
C = specific heat capacity of water = 4.184 J/g-°C
ΔT = Increase in temperature
ΔT for water = 43.4 - 21.4 = 22°C
= m × 4.184 × 22 …. (2)
Now
=
×ΔT
where = Heat capacity of pellet = 56J/°C
Δ T for pellet = 43.4 - 113 =- 69.6°C
= 56 × -69.6 = -3897.6 J
From equation (1) and (2)
-m× 4.184 × 22 =-3897.6
m= 42.34 g
Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.
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