Question

A 0.519 g sample of steam at 105.5 ∘C is condensed into a container with 5.91 g of water at 16.7 ∘C. What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18 J g⋅ ∘C, the specific heat of steam is 2.01 J g⋅ ∘C, and Δvap=40.7 kJ/mol.

Answer 1

The final temperature of the mixture is 28.93°C

Data;

• mass of sample $m_1 = 0.519g$
• initial temperature of solid $T_1 = 105.5^0C$
• mass of water $m_2 = 5.91g$
• Temperature of water $T_2 = 16.7^0C$
• Final Temperature$T_3 = ?$
• specific heat capacity of steam$c_1 = 2.01 J/g^0C$
• specific hear capacity of water $c_2 = 4.18 J/^0C$

Heat Transfer

The underlying principle of transfer of heat is discussed in heat transfer which can be summarized as heat lost by a warmer body is equal to heat gain by a colder body. The formula of this can be written as

$Q=mc\delta \theta$

Heat Loss = Heat gain

$m_1c_1(T_1-T_3)=m_2c_2(T_2-T_3)\\0.519*2.01*(105-T_3)=5.91*4.18*(16.7-T_3)\\109.53495-1.04319T_3=412.55346-24.7038T_3)\\1.04319T_3+24.7038T_3=412.55346-109.53495\\25.74699T_3=303.01851\\T_3=\frac{303.01851}{25.74699}\\T_3=11.77^0C$

The final temperature is equal to

$T_f = \delta vap - T_3\\T_f = 40.7 - 11.77\\T_f = 28.93^0C$

The final temperature of the mixture is 28.93°C

Learn more on heat transfer here;

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