What is the specific heat of the metal? See information below   a 50.0 sample of metal beads wass heated 100.0 C. the metal was added to 25.0 g of water at a temperature of 21.0C - temperature rise for water 40.0C.  use correct significant figures what was the temperature of the metal at equilibrium  what was the temperature change of the metal what was the temperature change of the metal   check_circle Expert Answer thumb_up   thumb_down Step 1 Given : Mass of metal used = 50.0 g Initial temperature of metal = 100.0 oC Initial temperature of water = 21.0 oC Final temperature of water = 40.0 oC And mass of water used = 25.0 g   Step 2 Since at equilibrium, the temperature of metal will be same as that of water. This is because, as per zeroth's law of thermodynamics, the temperature of 2 bodies present in contact will be same at equilibrium. Hence the temperature of metal at equilibrium = final temperature of water = 40.0 oC. The temperature change of the metal is given by  => temperature change of metal = final equilibrium temperature of metal - Initial temperature of metal  => temperature change of metal = 40.0 - 100.0 = -60.0 oC And the temperature change of water = final equilibrium temperature of water - Initial temperature of water => the temperature change of water = 40.0 - 21.0 = 19.0 oC

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What is the specific heat of the metal? See information below

 

a 50.0 sample of metal beads wass heated 100.0 C. the metal was added to 25.0 g of water at a temperature of 21.0C - temperature rise for water 40.0C. 

use correct significant figures

what was the temperature of the metal at equilibrium 

what was the temperature change of the metal

what was the temperature change of the metal

 

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Expert Answer

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Step 1

Given : Mass of metal used = 50.0 g

Initial temperature of metal = 100.0 oC

Initial temperature of water = 21.0 oC

Final temperature of water = 40.0 oC

And mass of water used = 25.0 g

 

Step 2

Since at equilibrium, the temperature of metal will be same as that of water.

This is because, as per zeroth's law of thermodynamics, the temperature of 2 bodies present in contact will be same at equilibrium.

Hence the temperature of metal at equilibrium = final temperature of water = 40.0 oC.

The temperature change of the metal is given by 

=> temperature change of metal = final equilibrium temperature of metal - Initial temperature of metal 

=> temperature change of metal = 40.0 - 100.0 = -60.0 oC

And the temperature change of water = final equilibrium temperature of water - Initial temperature of water

=> the temperature change of water = 40.0 - 21.0 = 19.0 oC

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