Question

Calculate the heat released when 25.0 grams of water freezes at 0 degree Celsius.

Answer 1

The total amount of heat released when 94.0 g of water at 80.0 °C cools to form ice at -30.0 °C is 75256 J.

Step-by-step explanation:

The total amount of heat released when 94.0 g of water at 80.0 °C cools to form ice at -30.0 °C can be calculated using the specific heat capacity of water and the heat of fusion of water. The formula to calculate the heat released is:

Q = mcΔT + mL

Where:

Q is the total heat released

m is the mass of water

c is the specific heat capacity of water

ΔT is the change in temperature

L is the heat of fusion of water

Using the values given:

m = 94.0 g

c = 4.18 J/g°C (specific heat capacity of water)

ΔT = 80.0 °C - (-30.0 °C) = 110.0 °C

L = 334 J/g (heat of fusion of water)

Substituting the values into the formula:

Q = (94.0 g)(4.18 J/g°C)(110.0 °C) + (94.0 g)(334 J/g)

Simplifying the equation gives:

Q = 43820 J + 31436 J = 75256 J

Therefore, the total amount of heat released when 94.0 g of water at 80.0 °C cools to form ice at -30.0 °C is 75256 J.

Answer 2

= 8350 joules

Solution and explanation:

• The heat of fusion refers to the quantity of heat released when a given amount of water freezes.

• For example, 1 g of water releases 334 J when it freezes at 0°C.

Therefore; For 25.0 g of water.

Heat released = Mass of water × heat of fusion

= 25 g × 334 J/g

= 8350 Joules

Hence, the amount of heat released when 25.0 g of water freezes at 0°C is 8350 J.