Constant-volume calorimeters are sometimes calibrated by running a combustion reaction of known δe and measuring the change in temperature. for example, the combustion energy of glucose is 15.57 kj/g. when a 1.500 g sample of glucose burns in a constant volume calorimeter, the calorimeter temperature increases from 21.45 to 23.34°c. find the total heat capacity of the calorimeter (in kj/k).

he total heat capacity of the calorimeter is - 0.09 kJ/K

Explanation:

Given information

the combustion energy of glucose E = 15.57 kJ/g

mass of glucose, m = 1.500 g

initial temperature, = 21.45

final temperature = 23.34

the heat capacity of the calorimeter can be determined by

C = Q/ΔT

where

C = heat capacity (kJ/)

Q = quantity of the heat (kJ)

ΔT = temeperature difference ()

Q = E m

   = - (15.57 kJ/g)(1.500 g)

   = 23.355 kJ

ΔT = -

     = 23.34 - 21.45

     = 1.89

Since the question ask to answer in kJ/k, we need to convert the temperature from celcius to kelvin by adding 273. thus,

ΔT = 1.89 + 273

     = 274.89 K

C = Q/ΔT

  = - 23.355 kJ/274.89 K

  = - 0.09 kJ/K

33: i think it’s b

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.