Home
Class 12
BIOLOGY
The rate of evaporation of water doubles...

 The rate of evaporation of water doubles with every rise in temperature by

A

(A) ` 5^(@)C`

B

(B) `100^(@)C`

C

(C) `1^(@)C`

D

(D) `10^(@)C`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the rate of evaporation of water and the temperature increase required for it to double, we can follow these steps: ### Step-by-Step Solution: 1. **Understand Evaporation**: - Evaporation is the process where water transforms from a liquid state to a vapor state. This occurs at the surface of the liquid. 2. **Identify the Forces at Play**: - Water molecules are held together by weak intermolecular forces (hydrogen bonds). To evaporate, these bonds need to be broken. 3. **Kinetic Energy Requirement**: - Water molecules possess kinetic energy, which helps in breaking these intermolecular forces. However, additional energy is required to facilitate this process. 4. **Role of Temperature**: - Temperature is a measure of the average kinetic energy of the molecules. Increasing the temperature provides more energy to the water molecules, increasing their kinetic energy. 5. **Experimental Rule**: - It has been experimentally determined that for every increase of 10 degrees Celsius in temperature, the rate of evaporation of water doubles. 6. **Conclusion**: - Therefore, the answer to the question is that the temperature must rise by **10 degrees Celsius** for the rate of evaporation to double. ### Final Answer: The rate of evaporation of water doubles with every rise in temperature by **10 degrees Celsius**. ---
Promotional Banner

Topper's Solved these Questions

  • TRANSPORT IN PLANTS

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE-E|10 Videos

  • TRANSPORT IN PLANTS

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE-C|10 Videos

  • THE LIVING WORLD

    VMC MODULES ENGLISH|Exercise CHAPTER EXERCISE-D|9 Videos

Similar Questions

Explore conceptually related problems

Evaporation of water is

In general, it is observed that the rate of a chemical reaction doubles with every 10° rise in temperature. If this generalisation holds for a reaction in the temperature range 298 K to 308 K, what would be the value of activation energy for this reaction ? (R = 8.314 JK^(-1) mol^(-1) )

The rate of reaction is doubled for every 10^(@)C rise in temperature. The increase in rate as result of an increase in temperature from 10^(@)C to 100^(@)C is how many times of the original rate?

Explain graphically how the rate of a reaction changes with every 10^@ C rise in temperature.

Select the correct statements. (i) Between temperature range 0 - 25^(@)C , rate of respiration doubles for every 10^(@)C rise in temperature. (ii) Cytochromes are iron-porphyrin compounds. (iii) Respiratory rate of wounded or injured plant parts generally decreases.

The rate of a reaction becomes 2 times for every 10^@C rise in temperature . How many times rate of reaction will be increased when temperature is increased from 30^@ C " to " 80^@C ?

The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every 10^(@)C rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every 10^(@)C change in temperature. "Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C) Arrhenius gave an equation which describes aret constant k as a function of temperature k = Ae^(-E_(a)//RT) where k is the rate constant, A is the frequency factor or pre-exponential factor, E_(a) is the activation energy, T is the temperature in kelvin, R is the universal gas constant. Equation when expressed in logarithmic form becomes log k = log A - (E_(a))/(2.303 RT) Activation energies of two reaction are E_(a) and E_(a)' with E_(a) gt E'_(a) . If the temperature of the reacting systems is increased form T_(1) to T_(2) ( k' is rate constant at higher temperature).