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Answer :
Final answer:
The total energy required to heat a 35.5g sample of ice from -17.5°C to liquid water at 77.3°C is 24612.5425 J, calculated by summing the energy to heat the ice, melt the ice, and heat the water.
Explanation:
To calculate the energy required to heat a 35.5g sample of ice from -17.5°C to liquid water at 77.3°C, we have to consider the energy needed to (1) heat the ice up to 0°C, (2) melt the ice, and (3) heat the resulting water up to 77.3°C:
Energy to heat the ice: q1 = mass of ice (g) × specific heat of ice (J/g°C) × temperature change (°C)
Energy to melt the ice: q2 = mass of ice (g) × heat of fusion of ice (J/g)
Energy to heat the water: q3 = mass of water (g) × specific heat of water (J/g°C) × temperature change (°C)
Here are the calculations:
q1 = 35.5g × 2.05 J/g°C × (0°C - (-17.5°C)) = 35.5g × 2.05 J/g°C × 17.5°C = 1268.4375 J
q2 = 35.5g × 333 J/g = 11820.5 J
q3 = 35.5g × 4.19 J/g°C × (77.3°C - 0°C) = 35.5g × 4.19 J/g°C × 77.3°C = 11523.605 J
The total energy required is the sum of q1, q2, and q3: Total Energy = q1 + q2 + q3 = 1268.4375 J + 11820.5 J + 11523.605 J = 24612.5425 J
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Rewritten by : Barada
Answer: Q = 6831.8 J
Explanation: To solve for Q we will use the heat capacity formula:
Q= mc∆T, where the specific heat capacity of ice is ( c= 2.03 J / g°C ).
Q= 35.5 g ( 2.03 J / g°C) ( 77.3°C - (-17.5°C) )
Q= 72.065 J/ °C ( 94.8 °C)
Q= 6831.8 J
