Calorimetry: How to Find Heat Lost?

Question

In Part 1 of the lab you will test how well a coffee cup calorimeter matches the ideal case. You will do this by adding a known mass of warmer than room temperature water (at a known temperature) to a known mass of water at room temperature in the calorimeter. The relevant equations are:

“heat lost” = (m_{hot water})(c_water)(ΔT_{hot water})

“heat gained” = (m_{room temp water})(c_water)(ΔT_{room temp water}) + q_calorimeter

where:

q_calorimeter = (calorimeter constant)(ΔT_calorimeter)

ΔT_calorimeter = ΔT_{room temp water}

As long as we use the same calorimeter, we do not need to know the mass of the calorimeter (the calorimeter constant is the product of the mass of the calorimeter and the heat capacity of the calorimeter). Ideally, the calorimeter constant would have a value of 0 J/°C. Consider the following data and answer the questions below.

When 50.0 mL of 60.0°C water was mixed in the calorimeter with 50.0 mL of 25.0°C water, the final temperature was measured as 41.1 °C. Assume the density for water is 1.000 g/mL regardless of temperature.

a. Determine the magnitude of the "heat lost" by the hot water. Do not enter a sign.

 J

Answer 1

heat lost by hot water = mass hot water x Cwater x ∆T = q = (50 g)(4.184 J/g/degree)(60 - 41.1)

q = 50 g x 4.184 J/g/deg x 18.9 degrees

q = 3954 J