ANSWER
[tex]\begin{equation*} 71.51\degree C \end{equation*}[/tex]
EXPLANATION
Parameters given:
Mass of thermometer, m = 300 g
Initial temperature of thermometer, t1 = 35°C
Volume of water, V = 258 cm³
Initial temperature of water, T1 = 80°C
First, let us find the mass of the water using the formula for density:
[tex]\begin{gathered} \rho=\frac{M}{V} \\ \Rightarrow M=\rho *V \end{gathered}[/tex]
where ρ = density of water = 1 g/cm³
Therefore, the mass of the water is:
[tex]M=1*258=258g[/tex]
According to the conservation of energy, the total heat flow (the sum of the heat energy of the thermometer and water) must be equal to 0 since no heat flows to the surroundings:
[tex]\begin{gathered} Q_g+Q_w=0 \\ mc(T-t_1)+MC(T-T_1)=0 \end{gathered}[/tex]
where c = specific heat capacity of glass thermometer = 0.2 cal/g°C
C = specific heat capacity of water = 1 cal/g°C
T = final temperature of thermometer and water
Hence, solving for T, we have that:
[tex]\begin{gathered} T=\frac{mct_1+MCT_1}{mc+MC} \\ T=\frac{(300*0.2*35)+(258*1*80)}{(300*0.2)+(258*1)} \\ T=\frac{2100+20640}{60+258}=\frac{22740}{318} \\ T=71.51\degree C \end{gathered}[/tex]
That is the final temperature of the thermometer.
