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Read and answer the question below
An 34.44-g sample of a metal is heated to 98.6oC in a hot water bath until thermal equilibrium is reached. The metal sample is quickly transferred to 50.0mL of water at 22.2oC contained in a calorimeter. The thermal equilibrium temperature of the metal sample plus water mixture is 28.3oC.


A. The change in temperature of the water is: ____ degrees. B. The change in temperature of the metal is:_____ degrees C. The specific heat of the metal is:0.527 J/(g*oC)

Respuesta :

scme1702
The change in water temperature is:
28.3 - 22.2
= 6.1 °C
The change in temperature of the metal is:
98.6 - 28.3
= 70.3 °C

Answer:

The change in water temperature is:

28.3 - 22.2

= 6.1 °C

The change in temperature of the metal is:

98.6 - 28.3

= 70.3 °C

Explanation: i just did it on edu.

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