Question

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

Answer 1

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

Answer 2

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

Step-by-step 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.

`Q=mc\Delta T`

`Q=35.0 g* 4.18 J/g^oC* 35.0^oC =5,120.5 J`

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