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PRE-LAB QUESTIONS: Name: Date: Partners Name: 1) A 19.29 g of unknown metal is heated to 99.7 C in a hot water bath. The quickly transferred to a calorimeter containing 500 ml of water at 23 The temperature for the mi (Density of water 1.0 g/mL) metal sample is final maximum xture, from the graph, is 26.4 °C. What is the specific heat of the metal? 2) During the experiment (finding specific heat of unknown metal), when hot metal is transferred to cold water, heat is inevitably lost. List all the different ways heat can be lost during the experiment Il such heat loss (question 2) increase/decrease no change the calculated value of the specific heat of the unknown metal? Explain your answer using equation 3) Wil 4) How can we make up for such heat loss in calculation? NaOH (HCl is the limiting reagent). The average initial temperature of both solutions is 21.6 °C. After the reaction, the highest recorded temperature is 24.4 °C. Calculate the heat of neutralization (kJrmol of HCl), [Density = 1.00 g/mL for both solutions, heat capacity = 418 J/gC for both solutions] 5) A chemist uses a coffee-cup calorimeter to neutralize 50.0 mL of 0.100 M HCl with 51.0 mL of CHE 112 -Lab: Coffee-Cup Calorimeter by Dr. Rajpara & Dr. lanniello 18/2016)


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Answer #1

Q1.

m = 19.29 g of metal

Tm = 99.7°C

V = 50 mL o fwater, T = 23.4°C

final Mx T = 26.4°C

find Cp metal.

The energy balance:

Qgain = -Qloss

Qwater= -Qmetal

Qwater = mwater * Cp water * (Tf-Twater)

Qmetal = m metal * Cp metal * (Tf-Tmetal)

then

mwater * Cp water * (Tf-Twater)= -m metal * Cp metal * (Tf-Tmetal)

substitute knonw data

mwater = Dwater*Vwater = 1 g/mL *50 mL = 50 g

50 *4.184 * (26.4-23.4)= -19.29 * Cp metal * (26.4-99.7)

Cp metal = 50 * 4.184 * (26.4-23.4) / (-19.29 *  (26.4-99.7) )

Cp metal = 0.4438 J/gC for this metla

Q2.

possible losses:

- Ambient

- Vaporization of water

- Sound/Work done by systems

- Loss due to container low/poor adiabatic status

Q3

Since Q is lost, then The temperature of the mtal will be lower, than the theoretical expected

if this is true, recall that

C metal = Q/(mass*dT)

if dT is lower, then, Cp metal is higher (incorrectly)

Q4

We can calculate the specific heat of the calorimeter in another experiment, this will be much precise

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