How much heat must be removed to change 75.0 grams of water at 25.0 C to ice at -35.0 C?
To determine the amount of heat that needs to be removed from 75.0 grams of water to change its temperature from 25.0°C to ice at -35.0°C, we need to consider two steps: 1) cooling the water from 25.0°C to 0°C, and 2) further cooling the water from 0°C to -35.0°C to convert it to ice.
Let's calculate the heat required for each step:
Step 1: Cooling water from 25.0°C to 0°C.
To find the heat required to cool the water to 0°C, we'll use the formula:
q = m * c * ΔT
where:
q is the heat required
m is the mass of the water (75.0 grams)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (from 25.0°C to 0°C)
Plugging in the values:
q₁ = 75.0 g * 4.18 J/g°C * (0°C - 25.0°C)
Step 2: Cooling water from 0°C to -35.0°C and turning it into ice.
To find the heat required to cool the water from 0°C to -35.0°C and convert it into ice, we'll use the formula:
q = m * ΔHf + m * c * ΔT
where:
q is the heat required
m is the mass of the water (75.0 grams)
ΔHf is the heat of fusion for water (334 J/g)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (from 0°C to -35.0°C)
Plugging in the values:
q₂ = 75.0 g * 334 J/g + 75.0 g * 4.18 J/g°C * (-35.0°C - 0°C)
Finally, to obtain the total heat required, we sum up the heat from both steps:
q_total = q₁ + q₂
By calculating q_total, we can find out how much heat must be removed to change 75.0 grams of water at 25.0°C to ice at -35.0°C.