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?

Question

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!

Answer 1

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.

Answer 2

This is how I did it:

(.034)(840)(21.6-38.7)+135(4186)(Ti-38.6)

Answer 3

To find the original temperature of the water, we need to use the concept of heat transfer.

First, we'll use the equation of heat transfer, which states that the heat gained by an object is equal to the mass of the object multiplied by its specific heat capacity multiplied by the change in temperature.

In this case, we have two objects: the glass thermometer and the water. We can set up an equation for the heat gained by each object and set the sum of these heats gained equal to zero, since no heat is gained or lost by the system as a whole.

Let's call the initial temperature of the water Ti.

The heat gained by the glass thermometer is given by:
massglass * specific heat capacity of glass * (21.6 - 38.7)

The heat gained by the water is given by:
masswater * specific heat capacity of water * (Ti - 38.7)

Setting the sum of these heats gained equal to zero, we have:

massglass * specific heat capacity of glass * (21.6 - 38.7) + masswater * specific heat capacity of water * (Ti - 38.7) = 0

Rearranging this equation, we can solve for the initial temperature (Ti):

masswater * specific heat capacity of water * (Ti - 38.7) = -massglass * specific heat capacity of glass * (21.6 - 38.7)

Solving this equation will give us the initial temperature (Ti) of the water.

Please note that the specific heat capacity of glass and water are known constants and can be looked up in reference material or provided in the problem statement.

I hope this helps! Let me know if you have any further questions.