Answer:
68.7 °C
Explanation:
You want the final temperature of a 300 g thermometer at 27 °C after it is put into 164 cm³ of water at 84 °C, assuming the system is isolated and the specific heats of the glass and water are 0.2 and 1.0 cal/(g·K), respectively.
The final temperature of the system is the weighted average of the initial temperatures, where the weights are the heat capacity of the glass and water.
The heat capacity of the glass thermometer is ...
(300 g)(0.2 cal/(g·K)) = 60 cal/K
The heat capacity of the water is ...
(164 cm³)(1.0 cal/(cm³·K)) = 164 cal/K
The final temperature will be ...
[tex]T_{\text{final}}=\dfrac{27\cdot60+84\cdot164}{60+164}=\dfrac{15396}{224}\approx68.7[/tex]
The final temperature of the thermometer is about 68.7 °C.
__
Additional comment
The units of cal/K cancel in the final computation, leaving only units of °C.
