Estimate the quantity of heat needed to melt 150 g of ice at $-10^{\circ} C$ to water at $15^{\circ} C$ [SHC of water $=4200 Jkg ^{-1} k ^{-1}$] [SHC of ice $=2100 Jkg ^{-1} k ^{-1}$ ], [SHC of ice $=3.36 \times 10^5 Jkg ^{-}$]
Answer (2)
Calculate the heat Q 1 to raise the ice temperature from -10°C to 0°C: Q 1 = m c i Δ T = 0.15 × 2100 × 10 = 3150 J .
Calculate the heat Q 2 to melt the ice at 0°C: Q 2 = m L f = 0.15 × 3.36 × 1 0 5 = 50400 J .
Calculate the heat Q 3 to raise the water temperature from 0°C to 15°C: Q 3 = m c w Δ T = 0.15 × 4200 × 15 = 9450 J .
Calculate the total heat required: Q t o t a l = Q 1 + Q 2 + Q 3 = 3150 + 50400 + 9450 = 63000 J . The quantity of heat needed is 63000 J .
Explanation
Problem Analysis We need to estimate the total heat required to melt 150g of ice at -10 degrees Celsius to water at 15 degrees Celsius. We will break this down into three steps: heating the ice to 0°C, melting the ice, and heating the water to 15°C.
Heating the Ice First, we calculate the heat Q 1 required to raise the temperature of the ice from -10°C to 0°C. We use the formula Q 1 = m c i \[ De lt a ] T , where m = 0.15 k g , c i = 2100 J k g − 1 K − 1 , and \[ De lt a ] T = 0 − ( − 10 ) = 10° C . Thus, Q 1 = 0.15 k g × 2100 J k g − 1 K − 1 × 10 K = 3150 J
Melting the Ice Next, we calculate the heat Q 2 required to melt the ice at 0°C to water at 0°C. We use the formula Q 2 = m L f , where m = 0.15 k g and L f = 3.36 × 1 0 5 J k g − 1 . Thus, Q 2 = 0.15 k g × 3.36 × 1 0 5 J k g − 1 = 50400 J
Heating the Water Then, we calculate the heat Q 3 required to raise the temperature of the water from 0°C to 15°C. We use the formula Q 3 = m c w \[ De lt a ] T , where m = 0.15 k g , c w = 4200 J k g − 1 K − 1 , and \[ De lt a ] T = 15 − 0 = 15° C . Thus, Q 3 = 0.15 k g × 4200 J k g − 1 K − 1 × 15 K = 9450 J
Total Heat Calculation Finally, we calculate the total heat required by summing the individual heat quantities: Q t o t a l = Q 1 + Q 2 + Q 3 = 3150 J + 50400 J + 9450 J = 63000 J
Final Answer Therefore, the total heat required to melt 150 g of ice at − 1 0 ∘ C to water at 1 5 ∘ C is 63000 J.
Examples
Imagine you're preparing a cold drink on a hot day. You start with ice from your freezer at -10°C and want to know how much energy it takes to turn that ice into refreshing 15°C water. This calculation helps you understand the energy needed for phase transitions and temperature changes, which is crucial in many applications, from cooking to industrial processes. By understanding these principles, you can optimize energy usage and predict outcomes in various thermal processes.
To melt 150 g of ice at -10°C to water at 15°C, we first heat the ice to 0°C (3150 J), then melt it (50400 J), and finally heat the water to 15°C (9450 J). The total heat required is 63000 J. Therefore, you need 63000 J of heat for this process.
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