The temperature of the water will reduce by 2.4 °C from 24.9 °C 22.5 °C.
The temperature change when 3.21 g of NH₄NO₃ is dissolved in 100 g of water under the same conditions is determined using the formula:
In example 5.6, when 3.21 g of NH₄NO₃ is dissolved in 50 g of water at 24.9°C in a calorimeter, the temperature decreases to 20.3°C.
The quantity of heat absorbed, q = -mcΔT
m = 50 + 3.21 = 53.21g
Assuming the specific heat of the solution is the same as that of water;
c = 4.18 J/g/K
ΔT = 20.3 - 24.9 = -4.6 K
q = -(53.21 * 4.18 * -4.6)
q = 1023.12 J
Since, the same mass of NH₄NO₃ is dissolved, the quantity of heat absorbed is the same.
ΔT = - q/mc
m = 100 + 3.21 = 103.21 g
ΔT = - 1023.12 J/103.21 * 4.18
ΔT = -2.4 °C
Therefore, the temperature of the water will reduce by 2.4 °C.
Learn more about specific heat of water at: https://brainly.com/question/26846829
#SPJ1
