Question

How many Joules of heat energy is required to convert 0.001 kg of ice at 0℃ to water at 45℃? Do NOT round, and do NOT include a unit in your answer.

LF = 3.33 x 105 J/Kg C = 4186 J/Kg℃

Answer 1

188.7 Joules.

Step-by-step explanation:

We can use the following equation to solve this:

Q = m * LF + m * C * ΔT

where:

• Q is the heat energy required (in Joules)
• m is the mass of the substance (in kilograms)
• LF is the latent heat of fusion (in Joules per kilogram)
• C is the specific heat capacity (in Joules per kilogram per degree Celsius)
• ΔT is the change in temperature (in degrees Celsius)

In this problem, we have the following information:

• m = 0.001 kg
• LF = 3.33 x 105 J/Kg
• C = 4186 J/Kg℃
• ΔT = 45℃

We can now plug these values into the equation to calculate the heat energy required:

Q = 0.001 kg * 3.33*105 J/Kg + 0.001 kg * 4186 J/Kg℃ * 45℃

Q=188.7 J

Therefore, the heat energy required is 188.7 Joules.