Explore a wide range of topics and get answers from experts on IDNLearn.com. Our platform offers reliable and detailed answers, ensuring you have the information you need.
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