A 34 g glass thermometer reads 21.6°C before it is placed in 135 mL of water. When the water and thermometer come to equilibrium, the thermometer reads 38.7°C. What was the original temperature of the water?
I need help on this please could someone please help out.Thank you!
Physics - bobpursley, Tuesday, April 24, 2012 at 5:53pm
the sum of heats gained is zero.
massglass*specheatglass*(21.6-38.7)+135*specheatwater*(Ti-38.6)=0
solve for Ti
Physics - Akbar, Tuesday, April 24, 2012 at 6:58pm
I got 15.43 degree Celsius but its wrong. I don't know why please help. Thank you!
physics - bobpursley, Wednesday, April 25, 2012 at 12:14pm
well, it is wrong, you know the final answer is between 21.6 and 38.7C
Let me see your calculations, including what you used as specific heats of glass and water.
physics - Akbar, Wednesday, April 25, 2012 at 12:36pm
This is how I did it:
(.034)(840)(21.6-38.7)+135(4186)(Ti-38.6)
m1•c1•(38.7-21.6) = m2•c2•(t-38.7)
m2 =ρ•V =1000• 135•10^-3•10^-3 =0.135 kg,
c1 =840 J/kg , c2 = 4180 J/kg.
t -38.7 = 0.034•840•17.1/0.135•4180 = 0.865 oC.
t = 38.7+ 0.865 =39.57oC
To solve this problem, you need to use the principle of heat equilibrium. The sum of heats gained by the glass thermometer and the water is equal to zero.
First, calculate the heat gained by the glass thermometer:
mass of the glass thermometer (m1) = 34 g
specific heat of glass (c1) = assuming it is 840 J/kg * °C
initial temperature of the glass thermometer (T1) = 21.6 °C
final temperature of the glass thermometer (T2) = 38.7 °C
Q1 = m1 * c1 * (T2 - T1)
Q1 = (34 g) * (840 J/kg * °C) * (38.7 °C - 21.6 °C)
Q1 = (34 * 0.001 kg) * (840 J/kg * °C) * (17.1 °C)
Q1 = 0.029 * 840 * 17.1 J
Q1 ≈ 411.444 J
Next, calculate the heat gained by the water:
mass of the water (m2) = 135 mL = 135 g
specific heat of water (c2) = 4186 J/kg * °C (assuming it is for liquid water)
initial temperature of the water (T2) = unknown
final temperature of the water and glass thermometer (T2) = 38.7 °C
Q2 = m2 * c2 * (T2 - T1)
Q2 = (135 g) * (4186 J/kg * °C) * (T2 - 38.7 °C)
Since heat gained by the glass thermometer is equal to the heat gained by the water, we can set up the equation:
Q1 + Q2 = 0
411.444 J + (135 g) * (4186 J/kg * °C) * (T2 - 38.7 °C) = 0
Now, you can solve for the unknown initial temperature of the water (T2):
411.444 J + (135 g) * (4186 J/kg * °C) * (T2 - 38.7 °C) = 0
Simplifying the equation:
411.444 J + (135 g) * (4186 J/kg * °C) * T2 - (135 g) * (4186 J/kg * °C) * 38.7 °C = 0
Rearranging the equation:
(135 g) * (4186 J/kg * °C) * T2 = (135 g) * (4186 J/kg * °C) * 38.7 °C - 411.444 J
Canceling out the units and isolating T2:
T2 = [(135 g) * (4186 J/kg * °C) * 38.7 °C - 411.444 J] / [(135 g) * (4186 J/kg * °C)]
Simplifying the expression:
T2 ≈ (5617290 - 411.444) / 563610
T2 ≈ 21.818 °C
Therefore, the original temperature of the water was approximately 21.818 °C.
To solve this problem, we need to use the principle of conservation of energy, which states that the heat gained by an object is equal to the heat lost by another object in a closed system.
In this case, the glass thermometer loses heat to the water, which gains heat. The sum of the heat gained and lost is zero.
We can express this as an equation:
mass_glass * specific_heat_glass * (initial_temperature - final_temperature) + mass_water * specific_heat_water * (initial_temperature_water - final_temperature_water) = 0
Let's now plug in the given values into this equation:
Mass of glass thermometer (m_glass) = 34 g
Specific heat of glass (C_glass) = ?
Initial temperature of the glass thermometer (T_initial) = 21.6°C
Final temperature of the glass thermometer (T_final) = 38.7°C
Mass of water (m_water) = 135 mL (which is equal to 135 g since the density of water is 1 g/mL)
Specific heat of water (C_water) = 4,186 J/(kg·°C)
Initial temperature of the water (T_initial_water) = ?
We have all the values except for the specific heat of glass. Unfortunately, specific heat values can vary depending on the type of glass. However, a safe assumption is that the specific heat of glass is similar to the specific heat of water, which is around 4,186 J/(kg·°C).
Now let's solve the equation for the initial temperature of the water (T_initial_water):
(34 g) * (specific heat of glass) * (21.6°C - 38.7°C) + (135 g) * (4,186 J/(kg·°C)) * (T_initial_water - 38.7°C) = 0
Solve the above equation for T_initial_water using the values given and the assumption for the specific heat of glass.
However, without the specific heat of glass, it is not possible to determine the exact value for the initial temperature of the water.