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.
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.© BrainMass Inc. brainmass.com July 22, 2018, 10:43 am ad1c9bdddf
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 ...
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.