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!
the sum of heats gained is zero.
massglass*specheatglass*(21.6-38.7)+135*specheatwater*(Ti-38.6)=0
solve for Ti
I got 15.43 degree Celsius but its wrong. I don't know why please help. Thank you!
To solve this problem, we can apply the principle of heat transfer. The heat gained by the thermometer is equal to the heat lost by the water.
First, let's determine the heat gained by the thermometer. We can use the equation:
Q = mcΔT
where:
Q is the heat gained or lost (in Joules),
m is the mass of the object (in grams),
c is the specific heat capacity (in J/g°C), and
ΔT is the change in temperature (in °C).
For the thermometer, the mass is given as 34 grams, and the change in temperature is (38.7 - 21.6) = 17.1°C. The specific heat capacity of glass is typically around 0.84 J/g°C.
So, the heat gained by the thermometer is:
Q = (34 g)(0.84 J/g°C)(17.1°C).
Next, let's determine the heat lost by the water. We can use the equation:
Q = mwcΔT
where:
Q is the heat gained or lost (in Joules),
mw is the mass of the water (in grams),
c is the specific heat capacity of water (in J/g°C), and
ΔT is the change in temperature (in °C).
For the water, the mass is given as 135 mL. However, we need to convert this to grams using the density of water, which is approximately 1 g/mL. So, the mass of the water is 135 grams. The specific heat capacity of water is approximately 4.18 J/g°C.
Now, we can solve for the original temperature of the water by equating the heat gained by the thermometer to the heat lost by the water:
(34 g)(0.84 J/g°C)(17.1°C) = (135 g)(4.18 J/g°C)(ΔT)
Solving for ΔT:
ΔT = (34 g)(0.84 J/g°C)(17.1°C) / (135 g)(4.18 J/g°C)
Calculating this, we find:
ΔT ≈ 0.852°C
Finally, we can find the original temperature of the water by subtracting ΔT from the final temperature of 38.7°C:
Original temperature = 38.7°C - 0.852°C
Calculating this, we find:
Original temperature ≈ 37.85°C
Therefore, the original temperature of the water was approximately 37.85°C.