A 300 g glass thermometer initially at 32 ◦C is put into 165 cm3 of hot water at 88 ◦C.
Find the final temperature of the ther- mometer, assuming no heat flows to the surroundings. The specific heat of glass is 0.2 cal/g ·◦ C and of water 1 cal/g ·◦ C.
Answer in units of ◦C.
To find the final temperature of the thermometer, we need to consider the heat transfer between the water and the thermometer.
The formula for heat transfer is given by:
Q = mcΔT
Where:
Q is the heat transferred.
m is the mass.
c is the specific heat.
ΔT is the change in temperature.
Let's start by finding the heat transferred from the water to the thermometer. Since no heat flows to the surroundings, the heat transferred from the water is equal to the heat absorbed by the thermometer.
The mass of the water is given as 165 cm^3. However, we need to convert it to grams since the specific heat is given in cal/g·°C.
Density of water = 1 g/cm^3
Mass of water = Density × Volume
Mass of water = 1 g/cm^3 × 165 cm^3
Mass of water = 165 g
Now, let's calculate the heat transferred from the water to the thermometer:
Q_water_to_thermometer = mcΔT
Q_water_to_thermometer = (165 g) × (1 cal/g·°C) × (T_final - 88°C) -- (Equation 1)
Next, let's find the heat absorbed by the thermometer. The mass of the thermometer is given as 300 g, and the specific heat of glass is given as 0.2 cal/g·°C.
Q_thermometer = mcΔT
Q_thermometer = (300 g) × (0.2 cal/g·°C) × (T_final - 32°C) -- (Equation 2)
Since the heat transferred from the water to the thermometer is equal to the heat absorbed by the thermometer, we can equate equations 1 and 2:
(165 g) × (1 cal/g·°C) × (T_final - 88°C) = (300 g) × (0.2 cal/g·°C) × (T_final - 32°C)
Now, let's solve this equation for T_final.
165 (T_final - 88) = 300 (0.2 T_final - 6.4)
165 T_final - 14520 = 60 T_final - 1920
105 T_final = 12600
T_final = 120°C
Therefore, the final temperature of the thermometer is 120°C.