How much heat in joules is released when 45.0 g of water (C = 4.184 J/gºC) is cooled from 25.0°C to 23.0°C?

– 376.56 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 45 g

Specific heat capacity (C) = 4.184 J/gºC

Initial temperature (T1) = 25 °C

Final temperature (T2) = 23 °C

Heat released (Q) =..?

Next, we shall determine the change in the temperature of the water. This can be obtained as follow:

Initial temperature (T1) = 25 °C

Final temperature (T2) = 23 °C

Change in temperature (ΔT) =.?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 23 – 25

Change in temperature (ΔT) = – 2 °C

Finally, we shall determine the heat released as follow:

Mass (M) = 45 g

Specific heat capacity (C) = 4.184 J/gºC

Change in temperature (ΔT) = – 2 °C

Heat released (Q) =..?

Q = MCΔT

Q = 45 × 4.184 × (– 2)

Q = – 376.56 J

Therefore, the heat released is – 376.56 J.

Note: The negative sign indicates that heat is released to the surrounding.

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.