A block of aluminum weighing 100g is cooled from 108.4°C to 68.2°C with the release of 1080 joules of heat. From this data, calculate the
specific heat of aluminum

Specific heat of aluminum  = 0.27 j/g.°C

Explanation:

Given data:

Mass of Al = 100 g

Initial temperature = 108.4°C

Final temperature = 68.2°C

Heat released = -1080 j

Specific heat of aluminum = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  68.2°C -  108.4°C

ΔT = - 40.2°C

Q = m.c. ΔT

-1080 j = 100 g ×c ×- 40.2°C

-1080 j = -4020 g.°C ×c

c = -1080 j/-4020 g.°C

c = 0.27 j/g.°C

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.