Final answer:
The amount of energy required to take a 15.0 g sample of ice at -15.0°C to liquid water at 45.0°C is approximately 8.3 kJ.
Step-by-step explanation:
The amount of energy required to take a 15.0 g sample of ice at -15.0°C to liquid water at 45.0°C can be calculated using the formula:
Q = mcΔT + mL
Where:
- Q is the total energy (in joules) required
- m is the mass of the sample (in grams)
- c is the specific heat capacity of ice (2.09 J/g°C)
- ΔT is the change in temperature
- L is the latent heat of fusion (334 J/g)
First, we need to calculate the energy required to heat the ice from -15.0°C to its melting point:
Q₁ = mcΔT = (15.0g)(2.09 J/g°C)(0 - (-15.0°C)) = 468.75 J
Next, we calculate the energy required to melt the ice:
Q₂ = mL = (15.0g)(334 J/g) = 5010 J
Finally, we calculate the energy required to heat the water from its melting point to 45.0°C:
Q₃ = mcΔT = (15.0g)(4.18 J/g°C)(45.0°C - 0) = 2827.5 J
Adding all the quantities of energy together:
Q = Q₁ + Q₂ + Q₃ = 468.75 J + 5010 J + 2827.5 J = 8306.25 J ≈ 8.3 kJ
Therefore, the amount of energy required is approximately 8.3 kJ. So, the correct answer is not listed among the given options.