Final answer:
The amount of heat required to change the sample from ice to liquid water involves multiple steps. First, the solid ice needs to be warmed, then melted, and finally, the liquid water needs to be warmed. Calculating the heat required for each step using the given values, the total heat required is 253. kJ.
Step-by-step explanation:
To calculate the amount of heat required to change the sample from ice at -45.0°C to liquid water at 75.0°C, we need to use the formula q = cmΔT. First, we need to calculate the heat required to warm the solid ice from -45.0°C to 0.00°C (q1). Using the specific heat capacity of ice (Cp, solid = 2.093 J/g⋅K), the mass of the water sample (100.0 g), and the change in temperature (ΔT = 0.00 - (-45.0) = 45.0 K), we get q1 = 100.0 g * 2.093 J/g⋅K * 45.0 K = 9391.85 J.
Next, we need to calculate the heat required to melt the solid ice (q2). Using the heat of fusion for ice (ΔHfusion = 40.7 kJ/mol), we can convert the mass of the water sample to moles (n) by dividing it by the molar mass of water (18.015 g/mol), and then multiply it by the heat of fusion: q2 = (100.0 g / 18.015 g/mol) * 40.7 kJ/mol = 226.3 kJ.
Finally, we need to calculate the heat required to warm the liquid water from 0.00°C to 75.0°C (q3). Using the specific heat capacity of liquid water (Cp, liquid = 4.184 J/g⋅K) and the change in temperature (ΔT = 75.0 - 0.00 = 75.0 K), we get q3 = 100.0 g * 4.184 J/g⋅K * 75.0 K = 31380.0 J.
The total amount of heat required (qtot) is the sum of q1, q2, and q3: qtot = 9391.85 J + 226.3 kJ + 31380.0 J = 252997.85 J = 253. kJ.