Six insulated containers hold 3,750 g of water at 24°C. A small copper cylinder is placed in each container; the masses and initial temperatures of the cylinders vary as given below. Rank the containers according to the maximum temperature of the water in each container after the cylinder is added, from largest to smallest. You may assume that the cylinder is completely submerged in the water. A. m = 250 g; T = 30°CB. m = 500 g; T = 60°CC. m = 750 g; T = 90°CD. m = 500 g; T = 15°CE. m = 750 g; T = 30°CF. m = 250 g; T = 60°C

the order from highest to lowest is      C> B> F> E> A> D

Explanation:

This is an exercise in calorimetry where the heat given off by the copper cylinder is equal to the heat absorbed by the water  

     Q_c = m_Cu ce_Cu ΔT_Cu  

    Q_a = m_water Ce_water ΔT  

    Q_c = Q_a  

    m_Cu ce_Cu (T_o-T_f) = m_water ce_water (T_f - T_i)

we clear the final temperature and substitute the values

With this expression we can know the final temperature of the system, let's substitute the values that are constant throughout the calculation

With this expression we can know the final temperature of the system, let's substitute the values that are constant throughout the calculation

a) m = 0.250kg, To = 30ºC  

      T_{fa}=(387 0.250 30 + 376740) / (15697.5 + 387 0.250)  

      T_{fa}= 379642.5 / 15794.25  

      T_{f} = 24.04ºC  

b) m = 0.500 kg, To = 60ºC  

     T_{fb} = (387 0.500 60 + 376740) / (15697.5 + 387 0.500)

     T_{fb} = 388350/15891  

     T_{fb = 24.44 ° C

c) m = 0.750 kg, To = 90ºC  

    T_{fc}= (387 0.750 90 + 376740) / (15697.5 + 387 0.750)  

    T_{fc}= 402862.5 / 15987.75  

    T_{fc} = 25.20ºC  

d) m = 0.500 kg, To = 15ºC  

    T_{fd} = (387 0.500 15 + 376740) / (15679.5 + 387 0.500)

    T_{fd} = 379642.5 / 15891  

    T_{fd} = 23.89ºC

the order from highest to lowest is      C> B> F> E> A> D

a straight line sloped