Answer to Question #135029 in Physical Chemistry for kaylee

Question #135029
A 92.9 piece of a silvery grey metal is heated to 178 degree celsius, and then quickly transferred into 75.0 mL of water initially at 24.0 degree celsius After 5 minutes both the metal and the water have reached the same temperature of 29.7 degrees celsius

what is the specific heat of the metal and identify the metal
1
Expert's answer
2020-09-29T06:23:46-0400

Solution.

"Q(H2O) = c(H2O) \\times \\rho(H2O) \\times V(H2O) \\times \\Delta T"

Q(H2O) = 1795.5 J

"c(Me) = \\frac{Q(H2O)}{m(Me) \\times \\Delta T1}"

"c(Me) = \\frac{1795.5}{0.0929 \\times (178-29.7)} = 130.33 \\frac{J}{kg \\times ^oC}"

Therefore, it is lead.

Answer:

"c(Me) = 130.33 \\frac{J}{kg \\times ^oC}"

Lead.


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