A thermometer of mass 0.0550 kg and of specific heat 0.837 kJ/kg•K reads 15.0°C. It is then completely immersed in 0.300 kg of water, and it comes to the same final temperature as the water. If the thermometer reads 44.0 °C, what was the temperature of the water before the insertion of the thermometer?
[mass therm x specific heat therm x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Solve for Ti H2O.
To find the initial temperature of the water before the insertion of the thermometer, we can use the principle of energy conservation.
First, let's calculate the heat absorbed by the thermometer when its temperature changed from 15.0°C to 44.0°C. We can use the formula:
q = mcΔT
Where:
q is the heat absorbed
m is the mass of the thermometer = 0.0550 kg
c is the specific heat of the thermometer = 0.837 kJ/kg•K
ΔT is the change in temperature = (44.0 - 15.0)°C
Substituting the values:
q = (0.0550 kg) * (0.837 kJ/kg•K) * (44.0°C - 15.0°C)
q = 0.1011 kJ
The heat absorbed by the thermometer is 0.1011 kJ.
Now, let's calculate the heat lost by the water when its temperature changed from the initial temperature to the final temperature (which is the same as the thermometer's temperature). We can use the same formula:
q = mcΔT
Where:
q is the heat lost by the water
m is the mass of the water = 0.300 kg
c is the specific heat of water = 4.18 kJ/kg•K (approximately)
ΔT is the change in temperature = (final temperature - initial temperature)
We know that the final temperature is 44.0°C, but we need to find the initial temperature. Let's call it T.
Therefore, the change in temperature for the water is:
ΔT = (44.0°C - T)
Substituting the values:
q = (0.300 kg) * (4.18 kJ/kg•K) * (44.0°C - T)
q = 1.2456 kJ * (44.0°C - T)
Now, according to the principle of energy conservation, the heat absorbed by the thermometer is equal to the heat lost by the water. Therefore:
0.1011 kJ = 1.2456 kJ * (44.0°C - T)
Now we can solve for T:
0.1011 kJ = 1.2456 kJ * (44.0°C - T)
0.1011 kJ / 1.2456 kJ = 44.0°C - T
0.0812 ≈ 44.0°C - T
T ≈ 44.0°C - 0.0812
T ≈ 43.92°C
Therefore, the initial temperature of the water before the insertion of the thermometer was approximately 43.92°C.