Question

Ans. - 1155 B6J 242 + 2024NO AH-3616T 21tr022420 1871.LRT 4. 100.0 mL of water is placed in a calorimeter and allowed to reach an initial temperature of 23°C. A 45.0g block of Cu (SH = 0.385 J/g."C) at 285°C is placed in the water (SH = 4.18 J/g. °C) in the calorimeter. a) If the final temperature reached is 30°C, determine the heat capacity of the calorimeter? b) The same calorimeter was used to determine the heat of solution of 20.0g of NH NO, NH4NO3 (► NH4NO3(aq) The initial temperature of the 100.0 mL of water and calorimeter were again 23°C. Once all of the NH4NO3 dissolved, the temperature reached was 11°C. Determine AH solution for NH4NO3. Ans. _J/mol

Answer 1

4-

In the given experiment, we are adding a hotter Cu-metal to a cooler water. Now when this happens, there is a heat transfer from the hotter body to cooler body until both the bodies come to the same temperature.

So the amount of heat released by the hotter body = amount of heat absorbed by cooler body = Q

This heat Q is calculated by the formula

Q = m * C * del T

where m = mass of the body

C = specific heat of the substance

del T = change in temp = T_final - T_initial

a -

Now given

Mass of Cu-metal = 45 g

Specific heat of Cu = 0.385 J/g^oC

T_initial = 285^oC

T_final = 30^oC

Thus del T = T_final - T_initial = 30^oC - 285^oC = -255^oC

So total amount of heat released by Cu is

Q = m * C * del T

Q = 45 g * 0.385 J/g^oC * (-255^oC)

= - 4417.875‬ J

Here the value of heat released = 4417.875‬ J, the -ve sign only indicates the loss of heat

Again for water,

Mass of waterl = 100 ml = 100 g

Specific heat of water = 4.18 J/g^oC

T_initial = 23^oC

T_final = 30^oC

Thus del T = T_final - T_initial = 30^oC - 23^oC = 7^oC

So total amount of heat absorbed by water is

Q = m * C * del T

Q = 100 g * 4.18 J/g^oC * (7^oC)

= 2926‬‬ J

That means the rest of the heat is absorbed by the caloriemeter

So heat absorbed by caloriemeter = 4417.875‬ J - 2926‬‬ J

= 1491.475‬ J

Again amount of heat absorbed by calorimeter can be calculated by the formula

Q = C×Δt

where C = specific heat of calorimeter

Δt = change in temp of calorimeter

putting the values-

Q = C×Δt

1491.475‬ J = C×7^oC

C = 1491.475‬ J / 7^oC

= 213.06 J/^oC

B-

Similarly for the given reaction NH4NO3(s) ----------> NH4NO3(aq)

Here Since the final temperature of both the water and calorimetere decreases, that means heat is lost by both of them

Given

Mass of waterl = 100 ml = 100 g

Specific heat of water = 4.18 J/g^oC

T_initial = 23^oC

T_final = 11^oC

Thus del T = T_final - T_initial = 11^oC - 23^oC = -12^oC

So total amount of heat lost by water is

Q = m * C * del T

Q = 100 g * 4.18 J/g^oC * (-12^oC)

= -5016‬‬‬ J

Similarly amount of heat lost by calorimeter can be calculated by the formula

Q = C×Δt

Q = 213.06 J/^oC ×(-12)^oC

= -2561.7‬‬ J

These heat lost by water and calorimeter must be absorbed by the reaction. So the heat absorbed by the reaction is

2561.7‬‬ J + 5016‬‬‬ J = 7577.7‬ J

Since the reaction absorbs heat, that means it is an Endothermic reaction. So the heat will be of +ve sign

Now the enthalpy of reaction = heat absorbed / moles of reactant

= heat absorbed / (mass of NH4NO3 taken / molar mass)

=  7577.7‬ J / (20 g / 80.043 g/mol)

= 7577.7‬ J / (0.249mol)

= 30432.53 J/mol