Help please!!

A 31.5g glass thermometer reads 26.5∘C before it is placed in 120ml of water. When the water and thermometer come to equilibrium, the thermometer reads 38.1∘C.

What was the original temperature of the water? [Hint: Ignore the mass of fluid inside the glass thermometer.]



TH2O = ∘C

Q = heat transfer = m*Cp*(Tf-Ti)
You know final temperature of thermometer and water, as well as initial temperature of the thermometer.
You have mass of the thermometer and you can use the density of water (nominally 1000 kg/m^3) to find the mass of water.
You can find the specific heat of both water and glass online or in your book.

The heat lost by the water from its initial temperature to the final temperature is the same heat used to bring the thermometer from its starting temperature to its final temperature.

Specific Heat of Glass: 0.84 J/gram/K
Specific Heat of Water: 4.179 J/gram/K
Density of Water: 1000 kg/m^3 = 1 kg / L = 1 g / mL
Mass of Water: 120g
Mass of Glass: 31.5g
Ti Glass: 26.5C
Tf Glass = Tf Water = 38.1C

Q_glass = -Q_water
(m_glass)(Cp_glass)(Tf_glass-Ti_glass) = (m_water)(Cp_water)(Ti_water-Tf_water)
(31.5)(0.84)(38.1-26.5) = (120)(4.179)(Ti_water-38.1)
(Ti_water-38.1) = 0.61206
Ti_water = 38.712C

This post was edited by Dontrunaway on Dec 2 2014 05:03pm

Thank you!