To calculate the total heat required to convert a 0.6 kg block of ice at -19°C to steam at 110.5°C, we need to consider several steps involving heating and phase changes. Here's the breakdown:

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Step 1: Heat the ice from -19°C to 0°C

The formula for heat transfer is: $q = mc\Delta T$ where:

• $m = 0.6 \, \text{kg}$ (mass of ice),
• $c_{\text{ice}} = 2.1 \, \text{kJ/kg/°C}$ (specific heat capacity of ice),
• $\Delta T = (0 - (-19)) = 19 \, \text{°C}$.

$q_1 = 0.6 \times 2.1 \times 19 = 23.94 \, \text{kJ}$

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Step 2: Melt the ice at 0°C to water

The heat required for phase change is: $q = mL_f$ where:

• $L_f = 334 \, \text{kJ/kg}$ (specific latent heat of fusion for ice).

$q_2 = 0.6 \times 334 = 200.4 \, \text{kJ}$

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Step 3: Heat the water from 0°C to 100°C

$q = mc\Delta T$ where:

• $c_{\text{water}} = 4.2 \, \text{kJ/kg/°C}$ (specific heat capacity of water),
• $\Delta T = 100 \, \text{°C}$.

$q_3 = 0.6 \times 4.2 \times 100 = 252 \, \text{kJ}$

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Step 4: Convert water at 100°C to steam at 100°C

$q = mL_v$ where:

• $L_v = 2260 \, \text{kJ/kg}$ (specific latent heat of vaporization for water).

$q_4 = 0.6 \times 2260 = 1356 \, \text{kJ}$

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Step 5: Heat the steam from 100°C to 110.5°C

$q = mc\Delta T$ where:

• $c_{\text{steam}} = 1.996 \, \text{kJ/kg/°C}$ (specific heat capacity of water vapor),
• $\Delta T = 110.5 - 100 = 10.5 \, \text{°C}$.

$q_5 = 0.6 \times 1.996 \times 10.5 = 12.56 \, \text{kJ}$

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Total Heat Required:

$Q_{\text{total}} = q_1 + q_2 + q_3 + q_4 + q_5$ $Q_{\text{total}} = 23.94 + 200.4 + 252 + 1356 + 12.56 = 1844.9 \, \text{kJ}$

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Final Answer:

The total heat required is 1844.9 kJ.

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Would you like a detailed breakdown of any specific step?

Related Questions:

1. What would happen if the initial temperature of the ice was even lower, say -40°C?
2. How does the specific heat capacity of water compare to that of ice and steam?
3. How would the calculation change if the mass of ice were doubled?
4. What physical phenomena account for the large latent heat values for fusion and vaporization?
5. How can this knowledge be applied to real-world energy calculations in saunas?

Tip:

Always check the units in thermal energy problems to avoid errors in conversions and ensure accuracy!