Question

What mass of water uses 7500 J to increase the temperature from 2 oC to 3 oC? The following information for water is given, but may or may not be useful: Δc = 4.184 J/goC ΔHfus = 334 J/g ΔHvap= 2260J/g

Answer 1

The mass of water that uses 7500 J to increase the temperature from 2°C to 3°C is 179.543 grams

Step-by-step explanation:

The given parameters are

Change in Heat , ΔH = 7500 J

The change in temperature ΔT = 2°C to 3°C = 3 - 1 = 1°C

The specific heat capacity for water is given as Δc = 4.184 J/(g·°C)

The heat of fusion,
`\Delta H_(fus)` = 334 J/g

The heat of vaporization,
`\Delta H_(vap)` = 2260 J/g

The formula for heat capacity is given as follows;

ΔH = m × Δc × ΔT

Given that the temperature of the process is above the melting point temperature of 0°C and below the boiling point temperature of 100°C, we make use only of the heat capacity of liquid water

Therefore, we have;

7500 J = m × 4.184 × 1

`m = (7500)/(4.184 * 1) = 1792.543 \ g`

The mass of water that uses 7500 J to increase the temperature from 2°C to 3°C = 179.543 grams.