When 100.0 g of an unknown metal at 98 degC were mixed in a perfect calorimeter

Question

When 100.0 g of an unknown metal at 98 degC were mixed in a perfect calorimeter with 50.0 g of water at 22 degC, the final temperature was observed to be 26.4 degC. If the unknown metal is one of those listed in below. Which one is it?

Substance                                     Specific Heat, J/g*degC

water                                                   4.184

aluminum                                            0.901

chromium                                            0.448

copper                                                 0.386

Iron                                                      0.450

Lead                                                    0.129

Nickel                                                  0.443

Tin                                                       0.217

Zinc                                                     0.386

Explanation / Answer

We know that M1C1 dT1= M2C2 d T2

mass of metal M1= 100g

Specifti heat of metal C1= x

dT1 of metal = Ti-Tf= 98-26.4=71.6

Mass of water M2= 50 g

Specific heat of water C2= 4.184

dT2 of water= Tf-Ti= 26.4-22=4.4

so substitute these values we get

100 * x * 71.6 = 50* 4.184 * 4.4

x= 0.129

so the metal is Lead

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