What is the final temperature in a squeezed cold pack that contains 48.5 g of nh4no3 dissolved in 125 ml of water? assume a specific heat of 4.18j/(g⋅∘c) for the solution, an initial temperature of 27.5 ∘c, and no heat transfer between the cold pack and the environment.

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

my dad did alot of meth and ace cheeks

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml