Calculate the amount of heat energy required to convert 35.0 g of ice at -12.5 $\degree$C to water at 24.0 $\degree$C.(C_water = 4.18 J/g $\degree$C; C_ice = 2.03 J/g $\degree$C; molar heat of fusion of ice = 6.01 x 10³ J/mol)

Question

Quizplus · Accepted Answer

To calculate the total heat energy, we need to consider three steps: heating the ice from -12.5°C to 0°C, melting the ice, and heating the resulting water from 0°C to 24.0°C. First, calculate the energy to heat the ice: q1 = m * C_ice * ΔT = 35.0 g * 2.03 J/g°C * 12.5°C = 888.75 J. Next, calculate the energy to melt the ice: q2 = n * ΔH_fusion = (35.0 g / 18.015 g/mol) * 6.01 x 10^3 J/mol = 11681.5 J. Finally, calculate the energy to heat the water: q3 = m * C_water * ΔT = 35.0 g * 4.18 J/g°C * 24.0°C = 3513.6 J. The total energy is q1 + q2 + q3 = 888.75 J + 11681.5 J + 3513.6 J = 16083.85 J, which is approximately 1.61 x 10^4 J.

Calculate the Amount of Heat Energy Required to Convert 35 °\degree°

Calculate the Amount of Heat Energy Required to Convert 35 $\degree$