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How much heat is required to increase the temperature of 35.0 grams of water from 10.0°C to 45.0°C? (The specific heat of water is 4.18 J/g°C) 4.18 J 6580 J 1460 J 5120 J

Respuesta :

imperialloyalist
You use the formula: Q=mc(Tfinal - Tinitial) where m is mass, c is specific heat 
Answer is 5120 J 

Answer:

Heat is required to increase the temperature of 35.0 grams of water is 5,120.5 J.

Explanation:

Mass of water = m = 35.0 g

Change in temperature of the water[=ΔT = 45.0°C to 10.0°C = 35.0°C

The specific heat of water = c = 4.18 J/g°C

Heat is required to increase the temperature of 35.0 grams of water be Q.

[tex]Q=mc\Delta T[/tex]

[tex]Q=35.0 g\times 4.18 J/g^oC\times 35.0^oC =5,120.5 J[/tex]

Heat is required to increase the temperature of 35.0 grams of water is 5,120.5 J.