ANSWER
![\begin{equation*} 71.51\degree C \end{equation*}](https://img.qammunity.org/2023/formulas/physics/college/mrmc2z1yw2ehcxcjcsj2263ze2y45v1qhj.png)
Step-by-step 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:
![\begin{gathered} \rho=(M)/(V) \\ \Rightarrow M=\rho *V \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/f2n9zoxqp0p0ahz6vp6zny5wkcl1zx3ngc.png)
where ρ = density of water = 1 g/cm³
Therefore, the mass of the water is:
![M=1*258=258g](https://img.qammunity.org/2023/formulas/physics/college/7bf94wnzxf5aah9khwkk3gs2izdygb6duu.png)
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:
![\begin{gathered} Q_g+Q_w=0 \\ mc(T-t_1)+MC(T-T_1)=0 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/rtlcwxqwiuhnee7igxb1ere1rhgk31a1d1.png)
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:
![\begin{gathered} T=(mct_1+MCT_1)/(mc+MC) \\ T=((300*0.2*35)+(258*1*80))/((300*0.2)+(258*1)) \\ T=(2100+20640)/(60+258)=(22740)/(318) \\ T=71.51\degree C \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/xbrku8e6hdbnw4vm87sdpd36oaztnlgwtb.png)
That is the final temperature of the thermometer.