Question

(AP CHEM) Thermochem Calorimeter problem: How to find final temperature given volume of water, mass of solid, and initial temp.? Question w/ answer given below↓ ↓ ↓ Need help on how to get the answer.

When 5g of NH4ClO4 is added to 100mL of water in a calorimeter, the temperature of the solution formed decreases by 3 degree Celsius. If 5g of NH4ClO4 is added to 1000mL of water in a calorimeter with initial temperature 25 degree Celsius, what is the final temperature approximately?

Ans: 24.7 degree Celsius.

Can someone show me how to get this answer?

Answer 1

There is an easy way to do this in less than 1 minute. Mass and volume are extensive, and the change in temperature is intensive.

So, 10 times more water would have 10 times the mass, which can absorb 10 times the heat. Therefore, since the same amount of heat was absorbed by the solid, the change in temperature must have been about 10 times as small.

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When `"5 g"` of `"NH"_4"ClO"_4` is added to `"100 mL"` of water, you say that `DeltaT = -3^@ "C"`.

For the same mass of `"NH"_4"ClO"_4` in `"1000 mL"` of water instead, the same amount of heat is absorbed into the `"NH"_4"ClO"_4` to break the lattice structure for its dissolution. Therefore:

`q_A = m_Ac_sDeltaT_A`
`q_B = m_Bc_sDeltaT_B`

The density of water is the same in both scenarios, and we can assume that they are approximately the same densities during the temperature change in both scenarios.

Therefore, `m_B/m_A ~~ V_B/V_A ~~ ("1000 mL")/("100 mL") = 10`,

where `V` is the volume and `m` is the mass. `c_s` is the specific heat of the solution, and `DeltaT` is the change in temperature.

From this, we find that since `q_A = q_B` due to the masses of `"NH"_4"ClO"_4` being the same and assuming conservation of energy,

`m_Ac_sDeltaT_A ~~ m_Bc_sDeltaT_B`

`(m_Acancel(c_s)DeltaT_A)/(m_Bcancel(c_s)DeltaT_B) ~~ 1`

Since `m_B/m_A ~~ 10`,

`(DeltaT_A)/(10DeltaT_B) ~~ 1`

Or, `color(blue)(DeltaT_B ~~ 1/10 DeltaT_A)`. Therefore, when you took 10 times the volume, you should see approximately 10 times smaller of a decrease in temperature. As a result, the final temperature is

`color(blue)(T_2) = 25^@ "C" - (3^@ "C")/10 = color(blue)(24.7^@ "C")`