A

Step 1: First, we need to find the amount of heat lost by the steam when it condenses and cools down to the final temperature \(T_f\). The steam will first condense to water at 373 K, releasing heat. The mass of steam is \(m_s = 0.05 \text{ kg} = 50 \text{ g}\).\
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Heat released during condensation: \(Q_{condensation} = m_s \times L_{vaporization} = 50 \text{ g} \times 540 \text{ cal/g} = 27000 \text{ cal}\)\
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Step 2: The condensed water will then cool down from 373 K to the final temperature \(T_f\). The specific heat of water is \(c_{water} = 1 \text{ cal/g K}\). The heat released by 50 g of water cooling from 373 K to \(T_f\) is: \(Q_{cooling} = m_s \times c_{water} \times (373 - T_f)\)\
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Step 3: Next, we find the heat gained by the ice to first melt and then warm up to the final temperature \(T_f\). The mass of ice is \(m_i = 0.45 \text{ kg} = 450 \text{ g}\). The heat required to melt the ice is: \(Q_{melting} = m_i \times L_{fusion} = 450 \text{ g} \times 80 \text{ cal/g} = 36000 \text{ cal}\)\
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After melting, the melted ice (water at 253 K) will warm from 253 K to \(T_f\): \(Q_{heating} = m_i \times c_{water} \times (T_f - 253)\)\
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Step 4: We now establish the energy balance equation: \Q_{condensation} + Q_{cooling} + Q_{melting} + Q_{heating} = 0\
27000 + 50(1)(373 - T_f) = -\(36000 + 450(1)(T_f - 253)\)
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Solving for \(T_f\): \
27000 + 50(373 - T_f) = -36000 - 450(T_f - 253)\
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Transposing and simplifying gives us a linear equation in terms of \(T_f\). By solving, we find that \(T_f = 328.5 \text{ K}\) rounded off to 329 K. This is the final temperature of the mixture. Therefore, the answer is 329 K.