Answer:
ΔT = - 2.13°C
Explanation:
Given data:
Mass of NH₄NO₃ = 48.5 g
Specific heat of solution = 4.18 j/g.°C
Initial temperature = 27.5°C
Final temperature = ?
Solution:
First of all we will find the moles of NH₄NO₃.
Number of moles = mass/molar mass
Number of moles = 48.5 g/80 g/mol
Number of moles = 0.6 mol
Now we will find the ΔH when we dissolve the 0.6 mol.
NH₄NO₃ + H₂O → NH₄NO₃ ΔH = +25.7 kJ
For 0.6 mol:
0.6 mol × +25.7 kJ/mol = 15.42 kj
15.42kj heat is absorbed by the reaction while -15.42 kj (-1542 j) heat will lost by the water.
The mass of water+ NH₄NO₃ = 125 g + 48.5 g
The mass of water+ NH₄NO₃ = 173.5 g
Q = m.c. ΔT
ΔT = T2 - T1
-1542 j = 173.5 g . 4.18 j/g.°C. ΔT
-1542 j = 725.23 j/°C. ΔT
ΔT = -1542 j / 725.23 j/°C
ΔT = - 2.13°C
Based on the data given, the final temperature in a squeezed cold pack is 6.2° C.
The moles of NH₄NO₃ is determined using the formula:
molar mass of NH₄NO₃ = 80 g/mol
Number of moles = 48.5/80
Number of moles = 0.6 moles
ΔH for the dissolution of 1 mole NH₄NO₃ = +25.7 kJ
ΔH for the dissolution of 0.6 mol = 0.6 × 25.7 kJ/mol
ΔH for the dissolution of 0.6 mol = 15.42 kj
Thus, 15.42 kJ heat is absorbed by the reaction while 15.42 kj heat will lost by the water.
Using the formula, Q = mcΔT to calculate the final temperature
mass of solution = 125 g + 48.5 g
mass of solution, m = 173.5 g
ΔT = T2 - T1
Q = 15.42 Kj = -15420 J
-15420 = 173.5 * 4.18 * ΔT
ΔT = -15420 / 725.23
ΔT = - 21.3° C
Final temperature, T2 = ΔT + T1
Final temperature, T2 = - 21.3° C + 27.5° C
Final temperature, T2 = 6.2° C
Therefore, the final temperature in a squeezed cold pack is 6.2° C.
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