The amount of heat required to change the sample from ice at -45.0°C to liquid water at 75.0°C is 83.8 kJ.
Quantity heat required to convert the ice to liquid
The total quantity of heat required is calculated as follows;
q(tot) = q1(to ice) + q2(fusion of ice) + q3(liquid)
q(tot) = mcΔθ₁ + mΔHf + mcΔθ₂
q(tot) = (100)(4.2)(0 - -45) + (334)(100) + (100)(4.2)(75 - 0)
q(tot) = 18,900 J + 33,400 J + 31,500 J
q(tot) = 18.900 kJ + 33.400 kJ + 31.500 kJ
q(tot) = 83,800 J = 83.8 kJ
Thus, the amount of heat required to change the sample from ice at -45.0°C to liquid water at 75.0°C is 83.8 kJ.
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