J.R. S. answered 03/29/20
Ph.D. University Professor with 10+ years Tutoring Experience
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
He B.
asked 03/29/20In 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” = (mhot water)(cwater)(ΔThot water)
“heat gained” = (mroom temp water)(cwater)(ΔTroom temp water) + qcalorimeter
where:
qcalorimeter = (calorimeter constant)(ΔTcalorimeter)
ΔTcalorimeter = ΔTroom 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.
J
J.R. S. answered 03/29/20
Ph.D. University Professor with 10+ years Tutoring Experience
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
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