Part A How much time (in minutes) will it take for the water heater to heat the 5.5 L of water from 23 ∘C to 43 ∘C? (Assume that the water has a density of 1.0 g/mL.)

Time = t = 30.01 minutes

Explanation:

The missing part of the question is that "The water heater power rating is 255 Watts, where 1 Watt = 1 J/s.

Firstly, calculating the work done by water heater to heat 5.5 L of water from 23° C to 43°C.

Work Done = Q = m C ΔT

where,

Q = Heat Transfer

m = mass of water

m = Density x Volume = 1 g/mL x 5.5 L = 1000g/L x 5.5L = 5500 g

ΔT = Temperature Difference = 43-23 = 20°C

C = Heat Capacity of Water = 4.186 J/g °C

Therefore,

Since, Power can be given as Work done per unit time,

or, in this case,

Here,

P = 255W or P = 255J (Since 1W = 1J)

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.

the answer is option a.