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Facebook YouTube Moodle Link for Chem124 prelab quiz md474enit..... Calorimetry Calculations Home Chegg.com 1. The specific heat of a metal is determined by heating the weighed metal, adding and measuring the it to water in a calorimeter Not yet answered temperature c The specific of water is 4-180 Joules per g per aC. The data collected are a igiiows: Marked out of 1.00 Flag question Grams of water in the calorimeter 43.13 Grams of metal 19.96 Initial temperature of metal 100.28 Initial temperature of water (C) 22.43 Maximum temperature of water after metal is added (C) 23.27 Calculate the following (1) Temperature fall of metal Answer: Question 2 1. The specific heat of a metal is determined by heating the weighed metal, adding it to water in a calorimeter, and measuring the a Not yet answered temperature change. The specific heat of water 4.180 Joules per g per The data collected are as follows: Marked out of 1.00 Flag question Grams of water in the calorimeter 43.13 Grams of metal 19.96 Initial temperature of metal 100.28 Initial temperature of water (oC) 22.43 Maximum temperature of water after metal is added (C) 2327 Calculate the following: Temperature rise of water Answer:

Explanation / Answer

In overall:

-Qmetal = Qwater

-mmetal * Cmetal * (Tf-Tmetal) = mwater * Cwater * (Tf-Twater)

-19.96 * Cmetal * (23.37-100.28) = 43.13* 4.184* (23.27-22.43)

Cmetla is required

Question 1.

for

Temperature fall, assume Tfinal of metal = Tfinal of water

so

dT metal = 23.27-100.28 = -77.01 °C

Question 2.

Temperature rise of water

dT = Tf water - Tinitial water = 23.27 - 22.43 = 0.84°C

Question 3.

Joules gained by water

Q = m*C(Tf-Tw)

Q = 43.13*4.184*(0.84) = 151.58 J

Question 4.

Specific heat of metal

-19.96 * Cmetal * (23.37-100.28) = 43.13* 4.184* (23.27-22.43)

-19.96 * Cmetal * (23.37-100.28) = 151.58

Cmetal = -151.58 / ( (23.37-100.28)*19.96 ) = 0.098741 J/gC

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