Estimate the quantity of heat needed to melt 150 g of ice at [tex]$-10^{\circ} C$[/tex] to water at [tex]$15^{\circ} C$[/tex].
[SHC of water [tex]$=4200 Jkg ^{-1} K^{-1}$[/tex] ]
[SHC of ice [tex]$=21001 kg^{-1} k ^{-1}$[/tex] ]
[SHC of ice [tex]$=3.36 \times 10^5 Jkg ^{-}$[/tex]]
Answer (1)
Calculate the heat Q 1 to raise the ice temperature from − 1 0 ∘ C to 0 ∘ C : Q 1 = m c i ce Δ T 1 = 0.15 × 2100 × 10 = 3150 J .
Calculate the heat Q 2 to melt the ice at 0 ∘ C : Q 2 = m L = 0.15 × 3.36 × 1 0 5 = 50400 J .
Calculate the heat Q 3 to raise the water temperature from 0 ∘ C to 1 5 ∘ C : Q 3 = m c w Δ T 2 = 0.15 × 4200 × 15 = 9450 J .
Calculate the total heat required: Q = Q 1 + Q 2 + Q 3 = 3150 + 50400 + 9450 = 63000 J .
Explanation
Problem Analysis We are asked to estimate the amount of heat needed to transform 150g of ice initially at − 1 0 ∘ C into water at 1 5 ∘ C . This process involves three steps: first, heating the ice to its melting point ( 0 ∘ C ); second, melting the ice into water at 0 ∘ C ; and third, heating the water to the final temperature of 1 5 ∘ C . We will calculate the heat required for each step and then sum them up to find the total heat needed.
Heating the Ice The first step is to calculate the heat Q 1 required to raise the temperature of the ice from − 1 0 ∘ C to 0 ∘ C . We use the formula: Q 1 = m c i ce Δ T 1 where m = 0.15 k g is the mass of the ice, c i ce = 2100 J k g − 1 K − 1 is the specific heat capacity of ice, and Δ T 1 = 0 − ( − 10 ) = 1 0 ∘ C is the change in temperature. Plugging in the values, we get: Q 1 = 0.15 k g × 2100 J k g − 1 K − 1 × 1 0 ∘ C = 3150 J
Melting the Ice The second step is to calculate the heat Q 2 required to melt the ice at 0 ∘ C into water at 0 ∘ C . We use the formula: Q 2 = m L where m = 0.15 k g is the mass of the ice and L = 3.36 × 1 0 5 J k g − 1 is the latent heat of fusion of ice. Plugging in the values, we get: Q 2 = 0.15 k g × 3.36 × 1 0 5 J k g − 1 = 50400 J
Heating the Water The third step is to calculate the heat Q 3 required to raise the temperature of the water from 0 ∘ C to 1 5 ∘ C . We use the formula: Q 3 = m c w Δ T 2 where m = 0.15 k g is the mass of the water, c w = 4200 J k g − 1 K − 1 is the specific heat capacity of water, and Δ T 2 = 15 − 0 = 1 5 ∘ C is the change in temperature. Plugging in the values, we get: Q 3 = 0.15 k g × 4200 J k g − 1 K − 1 × 1 5 ∘ C = 9450 J
Total Heat Calculation Finally, we calculate the total heat Q required by summing the heat from each step: Q = Q 1 + Q 2 + Q 3 = 3150 J + 50400 J + 9450 J = 63000 J Therefore, the total heat required to melt 150 g of ice at − 1 0 ∘ C to water at 1 5 ∘ C is 63000 J.
Final Answer The estimated quantity of heat needed to melt 150 g of ice at − 1 0 ∘ C to water at 1 5 ∘ C is 63000 J .
Examples
This calculation is crucial in many real-world applications, such as designing efficient cooling systems, predicting ice melting rates in climate models, and optimizing food storage techniques. For instance, engineers use these principles to design refrigerators that can effectively cool food items, considering the heat required to lower the temperature and potentially freeze water content in the food. Similarly, climate scientists use these calculations to estimate the energy needed to melt ice sheets, which helps in understanding and predicting sea-level rise due to global warming. Even in culinary arts, understanding heat transfer is essential for processes like ice cream making, where controlling the freezing and melting rates is vital for achieving the desired texture.
