Question

1) Use the simulator to react 51 mL of 2.6 M HNO2 with 94 mL of 3.0 M NaOH; note the initial and final temperature. Use the data from the simulation and estimate the density of the liquid to be 1.00 g/mL to calculate the change in enthalpy per mole of product (n). Express your answer in kilojoules per mole to three significant figures.

2) A mass of 33.5 g of an unknown solid initially at 140.0 ∘C is added to an ideal constant pressure calorimeter containing 100.0 g of water (Cs,water=4.184J/(g⋅∘C)) initially at 20.0 ∘C. After the mixture reaches thermal equilibrium, the final temperature is recorded to be 34.19 ∘C. What is the specific heat capacity of the unknown solid?

3) A new metal alloy is found to have a specific heat capacity of 0.548  J/(g⋅∘C). First, 22.5 g of the new alloy is heated to 170. ∘C. Then, it is placed in an ideal constant pressure calorimeter containing 90.0 g of water (Cs,water=4.184 J/(g⋅∘C)) at an initial temperature of 20.0 ∘C. What will the final temperature of the mixture be after it attains thermal equilibrium?

Answer 1

Answer 1. You need to perform the experiment and need initial/final temperature.

Answer 2.

Given that,

Initial Temperature of water = 20 ^oC

Initial Temperature of solid = 140 ^oC

Final Temperature of mixture = 34.19 ^oC

Mass of solid = 33.5 g

Mass of water = 100 g

Specific heat capacity of solid = k

Specific heat capacity of water (s) = 4.184 J/g-^oC

According to principle of calorimeter, (total heat = 0), hence

Heat lost by solid = Heat gained by water

m₁k = m₂s

33.5*k*(140-34.19) = 100*4.184*(34.19-20)

k = 1.675 J/g-^oC

Therefore,

Specific Heat of Solid = 1.675 J/g-^oC.

Answer 3.

Given that,

Initial Temperature of water = 20 ^oC

Initial Temperature of solid = 170 ^oC

Final Temperature of mixture = T ^oC

Mass of solid = 22.5 g

Mass of water = 90 g

Specific heat capacity of solid (k) = .548 J/g-^oC

Specific heat capacity of water (s) = 4.184 J/g-^oC

According to principle of calorimeter, (total heat = 0), hence

Heat lost by solid = Heat gained by water

m₁k = m₂s

22.5*0.548*(170-T) = 90*4.184*(T-20)

(170-T) = 30.54*(T-20)

31.54 T = 30.54*20 + 170

T = 24.76 ^oC

Therefore,

Final temperature of Mixture = 24.76 ^oC