Experiment 14 water in a calo- tal sample weighing 147.90 g and at a temperature

Question

Experiment 14 water in a calo- tal sample weighing 147.90 g and at a temperature of 99.5 C was placed in 49.73 g of nimeter at 23.0 C. At equilibrium the temperature of the water and metal was 41.8 C a. What was ar for the water? (A-1 Advance Study Assignment: Heat Effects and Calorimetry I. oC b. What was for the metal? oC c. How much heat flowed into the water? (Take the specific heat of the water to be 4.18 J/g 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 when 498 g of NaOH was dissolved in 4972 g of water in a calorimeter at 23.7°C, the temperature of the solution went up to 50.1°C. a. Is this dissolution reaction exothermic?Why? 2 b. Calculate go, using EquationI joules meter (Eq. 5) c. Find AH for the reaction as it occurred in the calori joules (continued on following page

Explanation / Answer

deltaT for water= final temperature of water-initial temperature = 41.8-23=18.8 deg.c

deltaT for metal ( metal looses heat)= 41.8-99.5= -57.7 deg.c

since the system is considered to be adiabatic

heat lost by metal = heat gained by water

-mass of metal* specific heat* temperature difference= mass of water* specific heat of water* change in temperature of water

Let CP= specific heat of metal given specific heat of water= 4.184

Heat flowed into water= 49.73*4.184*18.8=3911.72 Joules

Heat lost by metal = -147.9*Cp*(-57.7) = 3911.72
Cp = 3911.72/(147.9*57.7)= 0.46 J/gm.deg.c

Based on the specific heat data, the metal is iron whose atomic weight is 56 g/mole.

2. Since there is a rise in temperature, the dissolution of NaOH in water is exothermic.

So heat taken by water= mass of water* specific heat of water* temperature difference

=49.72*4.184*(50.1-23.7)=5492 joules

deltaH= -total mass * specific heat of solution* temperature change

specific heat of solution is taken as specific heat of water

=-(4.98+49.72)*4.184* (50.1-23.7)=-6042

Moles of NaOH= mass of NaOH/Molar mass = 4.98/40=0.1245

Enthalpy change/mole of NaOH= -6042/0.1245 J/mole =-48530.4 J/mole= -48.530 Kj/mole

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