Respuesta :
The final temperature of the mixture is 28.93°C
Data;
- mass of sample [tex]m_1 = 0.519g[/tex]
- initial temperature of solid [tex]T_1 = 105.5^0C[/tex]
- mass of water [tex]m_2 = 5.91g[/tex]
- Temperature of water [tex]T_2 = 16.7^0C[/tex]
- Final Temperature[tex]T_3 = ?[/tex]
- specific heat capacity of steam[tex]c_1 = 2.01 J/g^0C[/tex]
- specific hear capacity of water [tex]c_2 = 4.18 J/^0C[/tex]
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
[tex]Q=mc\delta \theta\\[/tex]
Heat Loss = Heat gain
[tex]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[/tex]
The final temperature is equal to
[tex]T_f = \delta vap - T_3\\T_f = 40.7 - 11.77\\T_f = 28.93^0C[/tex]
The final temperature of the mixture is 28.93°C
Learn more on heat transfer here;
https://brainly.com/question/13439286
