Advance Study Assignment: Heat Effects and Calorimetry metal sample weighing 147

Question

Advance Study Assignment: Heat Effects and Calorimetry metal sample weighing 147.90 g and at a temperature of 99.5"C was placed in 49.73 g of water in a calo- rimeter at 23.0°C. At equilibrium the temperature of the water and metal was 41.8C a. what was ar for the water? (Ar-t -t-) b. What was Ar for the metal? °C c. How much heat flowed into the water? (Take the specific heat of the water to be 4.18 1/ig C) _joules d. Calculate the specific heat of the metal, using Equation 3. joules/g"C e. What is the approximate molar mass of the metal? (Use Eq.4.) g/mol 2. When 4.98 g of NaOH was dissolved in 49,.72 g of water in a calorimeter at 23.7 'C, the temperature of the solution went up to 50.1°C. Why? a. Is this dissolution reaction exothermic? b. Calculate Qho using Equation 1. joules c. Find AH for the reaction as it occurred in the calorimeter (Eq. 5).

Explanation / Answer

1.

Given:

Mass of metal sample: 147.90 g

Initial temperature of Metal: 99.5 °C

Final temperature of Metal: 41.8 °C

Mass of water: 49.73 g

Initial temperature of Water: 23.0 °C

Final temperature of Water:? 41.8 °C

a. What was the ?T for water?

?T (water) = Final temp of water - initial temp of water = 41.8 °C - 23.0 °C = 18.8 °C

b. What was the ?T for Metal?

?T (Metal) = Final temp of Metal - initial temp of Metal = 41.8 °C - 99.5 °C = -57.7 °C

C. How much heat flowed into the water?

Heat flowed in to water can be calculated using the formula:

q = mCs?Twater

q = 49.73 g * 4.18 J/g°C * 18.8 °C = 3908J = 3.91KJ

d. Calcualate the specific heat of metal?

Heat flowed in to water = Heat lost by metal

Therefore, heat lost by metal = - 3098 J (-ve symbol indicating the loss)

qmetal = mCs?Tmetal

- 3098 J = 147.90 g * Cs(metal) * -57.7 °C  

Cs(metal) = (- 3098 J )/(147.90 g * -57.7 °C)

Cs(metal) = 0.458 J/g °C

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