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An object with a specific heat of 0.800 J/gºC at a temperature of 405 ºC is...

An object with a specific heat of 0.800 J/gºC at a temperature of 405 ºC is placed in a calorimeter with a heat capacity of 2.00 J/ºC that contains 100.0 mL of water. The initial temperature of the water and the calorimeter is 25.3ºC. After the object is added to the calorimeter, the final temperature of the calorimeter, water, and object is 29.8ºC. What is the mass of the object in grams?

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Answer #1

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kg∙K), ∆ is a symbol meaning "the change in"

∆T = change in temperature (Kelvins, K)

Heat lost by the object = heat gained by water + Calorimeter constant

m g x 0.8 J / g 0C x (405 - 29.8) 0C = 100 gm x 4.184  J/gºC x (29.8 - 25.3) 0C + (405 - 29.8)ºC x 2 J/ºC

300.16 m = 1882.8 + 750.4 Joules

300.16m = 2633.2 Joules

m = 8.77 gm

Mass of the objective is 8.77 gm

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