How many grams of water would require 2.20 x 10 4joules of heat to raise its temperature from 34.0°C to 100.0°C? The specific heat of water is 4.18 J/g∙°C

1 answer

To determine the grams of water that would require 2.20 x 10^4 joules of heat, we need to calculate the amount of heat absorbed by the water during the temperature change.

The equation to calculate the amount of heat absorbed is:

q = m * c * ΔT

where:
q = amount of heat absorbed
m = mass of the water (grams) - unknown
c = specific heat of water (4.18 J/g∙°C)
ΔT = change in temperature (final temperature - initial temperature)

Substituting the given values into the equation:

2.20 x 10^4 J = m * (4.18 J/g∙°C) * (100.0°C - 34.0°C)

Simplifying:

2.20 x 10^4 J = m * 4.18 J/g∙°C * 66.0°C

Dividing both sides of the equation by (4.18 J/g∙°C * 66.0°C):

2.20 x 10^4 J / (4.18 J/g∙°C * 66.0°C) = m

Simplifying:

m = 79.686 g

Therefore, it would require 79.686 grams of water to raise its temperature from 34.0°C to 100.0°C.