Question

3. A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the temperature of the water in calorimeter is determined to be 19.1 °C. To this is added 50.0 mL of water that was originally a temperature of 54.9 °C. A careful plot of the recorded temperature established T0 as 31.3 °C. What is the calorimeter constant (J/°C)?

Answer 1

The calorimeter constant is 816.800 J/degree.

Step-by-step explanation:

Data given:

volume of cold water = 50 ml

initial temperature of the cold water t1 = 19.1 degrees

volume of hot water = 50 ml

temperature of hot water t2 = 54.9 degrees

change in temperature t3= 31.3 degrees

calorimeter constant = ?

energy gained by cold water = energy lost by hot water

q hot water = 50 x 4.184 x (54.9-31.3)

q hot water = 50 x 4.184 x 23.6

= 4937.9 cal/degree

q cold water = 50 x 4.184 x (31.3-19.1)

q cold water = 2552.24 cal/deg

FINAL ENERGY = qhot water - q cold water

final energy = 4937.9 - 2552.24

final energy = 2385.66 cal/deg

calorimeter constant =
`(final energy)/(t3-t1)`

putting the values:

calorimeter constant =
`(2385.66)/(12.2)`

CALORIMETER CONSTANT = 195.22 J/deg

to convert in joules/degree multiply the value with 4.184

= 195.22 x 4.184

= 816.800 J/deg