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?
The temperature of the water will drop as heat is added to the thermometer as it reaches an equilibrium temperature.
The specific heat of the water is 4.18 kJ/Kg*K
heat gained by thermometer
= heat lost by water
0.055*(0.837)(44 - 15) = 0.300*(4.18)(T - 44)
1.335 = 1.254 (T-44)
T - 44 = 1.08
T = 45.08 C is the initial water temperature
To solve this problem, we need to use the principle of energy conservation.
The amount of heat gained by the thermometer will be equal to the amount of heat lost by the water. We can use the equation:
Q_t = Q_w
where:
Q_t = heat gained by the thermometer
Q_w = heat lost by the water
We can calculate the heat gained by the thermometer using the formula:
Q_t = (mass_t)(specific heat_t)(change in temperature_t)
where:
mass_t = mass of the thermometer
specific heat_t = specific heat of the thermometer
change in temperature_t = final temperature - initial temperature of the thermometer
Substituting the given values, we have:
Q_t = (0.0550 kg)(0.837 kJ/kg•K)(44.0°C - 15.0°C)
Simplifying the expression, we find:
Q_t = (0.0550 kg)(0.837 kJ/kg•K)(29.0°C)
Q_t = 0.13199 kJ
Now, let's calculate the heat lost by the water using the formula:
Q_w = (mass_w)(specific heat_w)(change in temperature_w)
where:
mass_w = mass of the water
specific heat_w = specific heat of the water
change in temperature_w = final temperature - initial temperature of the water
We need to find the initial temperature of the water, so we rearrange the equation:
initial temperature of the water = final temperature - (Q_w / (mass_w)(specific heat_w))
Substituting the given values, we have:
initial temperature of the water = 44.0°C - (0.13199 kJ / (0.300 kg)(4.184 kJ/kg•K))
Simplifying the expression, we find:
initial temperature of the water ≈ 44.0°C - 0.02936°C
Therefore, the initial temperature of the water was approximately 44.0°C - 0.02936°C = 43.97064°C (rounded to 5 decimal places).
Hence, the temperature of the water before the insertion of the thermometer was approximately 43.97064°C.
To solve this problem, we need to apply the principle of conservation of energy.
Let's denote:
- m1 as the mass of the thermometer (0.0550 kg),
- c1 as the specific heat capacity of the thermometer (0.837 kJ/kg•K),
- T1 as the initial temperature of the thermometer (15.0°C),
- m2 as the mass of the water (0.300 kg),
- c2 as the specific heat capacity of water (4.186 kJ/kg•K),
- T2 as the initial temperature of the water (unknown),
- Tf as the final temperature of both the thermometer and the water (44.0°C).
We know that the change in internal energy of an object can be calculated using the equation:
ΔQ = m * c * ΔT
Where:
ΔQ is the change in internal energy,
m is the mass of the object,
c is the specific heat capacity of the object,
ΔT is the change in temperature.
First, let's calculate the change in internal energy of the thermometer (ΔQ1):
ΔQ1 = m1 * c1 * (Tf - T1)
Next, let's calculate the change in internal energy of the water (ΔQ2):
ΔQ2 = m2 * c2 * (Tf - T2)
Since the total change in internal energy is conserved, ΔQ1 = -ΔQ2.
Therefore, we can write:
m1 * c1 * (Tf - T1) = -m2 * c2 * (Tf - T2)
Let's plug in the given values:
0.0550 kg * 0.837 kJ/kg•K * (44.0°C - 15.0°C) = -0.300 kg * 4.186 kJ/kg•K * (44.0°C - T2)
Now, let's solve for T2:
0.0550 kg * 0.837 kJ/kg•K * 29.0°C = 0.300 kg * 4.186 kJ/kg•K * (44.0°C - T2)
By simplifying the equation, we get:
0.0008705 kJ * (44.0°C - 15.0°C) = 1.25538 kJ * (44.0°C - T2)
Now, let's solve for T2:
(44.0°C - T2) = (0.0008705 kJ * (44.0°C - 15.0°C)) / 1.25538 kJ
T2 = 44.0°C - [(0.0008705 kJ * (44.0°C - 15.0°C)) / 1.25538 kJ]
After doing the calculations, we find T2 to be approximately 33.2°C.
Therefore, the temperature of the water before the insertion of the thermometer was approximately 33.2°C.