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Specific Heat & Enthalpy

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Question #1

A 25.000g sample of unknown metal is heated to 99.5 degrees Celsius and added to 50.0mL of water in a calorimeter, which has an initial temperature of 22.o degrees Celsius (density of water is 0.99780 g/mL. The temperature of the calorimeter increases to a maximum temperature of 33.5 degrees Celsius. The heat capacity of the calorimeter is 19.0J/degrees Celsius. The specific heat of water is 4.184J/gram degrees Celsius.Calculate the specific heat of the unknown metal.

Question #2

A 8.375g sample of NaNO3 (F.W. = 85 amu) is added to a calorimeter with 100.0mL of water at 23.0 degrees celcius (density of water = 0.99756 g/mL). The temperature drops to 14.8 degrees Celsius. The heat capacity of the calorimeter is 19.0J/degrees Celsius and the specific heat of NaNO3 is 3.85 J/g degrees Celsius. Calculate the enthalpy of solution and the molar enthalpy of solution.

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Solution Summary

In this solution we consider two different questions regarding the use of specific heat. In the first, we use experimental data to figure out the specific heat of a metal that is unknown. In the second, we figure out how to use specific heat to calculate enthalpy of a solution.

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25.000g of unknown metal
50.0 mL water
Initial temp of metal - 99.5 celsius
Initial temp of water - 22.0 celsius
Final temperature of water (and metal) - 33.5 celsius

Calculate energy that was absorbed by the water

q = (0.99780g/mL * 50.0mL) * 4.184J/gC * (33.5 - 22.0)
q = 2.40 x 10^3 J absorbed, which is ...

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