Water in both the containers is heated with same heat source. It means, the rate of heat addition in both the containers is same. Water in container A is to be heated from 10^{0}C to 20^{0}C. Water in container B is to be heated from 20^{0}C to 30^{0}C.

#### Question

With above conditions, answer the following question:

To increase the temperature of water (10^{0}C to 20^{0}C for container A, and 20^{0}C to 30^{0}C for container B), time required for heating will be

- Same for both the containers
- Container A will take more time than container B
- Container B will take more time than container A
- Insufficient data provided. Cannot be predicted.

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**Best answer**

The best answer selected is given by “Bruno”. Bruno writes:

*“Container A will take more time then container B.**The time would only be the same if we could consider that the enthalpy variation was linear, which it isn't (although it almost is). But the enthalpy change from 20 to 30 C is slightly smaller than from 10 to 20, so heating container B from 20 to 30 should take a slightly smaller time.”*

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#### Answer by LearnCAx Team

The correct answer is

2: Container A will take more time than container B

**Importance of assumptions**

It is assumed that there is no heat loss from container to atmosphere. If we do not make this assumption, container B will have more heat loss as compared to container A. Without this assumption, this difference will play some role in deciding heating time.

It is assumed that rate of heat addition in both containers is same. Note that we are assuming the amount of heat addition is same. In case even if heat source is same, the rate of heat addition is decided by conductivity, viscosity (as this is natural convection). As we are assuming that rate of heat addition is same, mean no matter what happens inside the tank, the water inside tank is going to absorb the same amount of heat. So conductivity and viscosity is not playing any role in time required to increase the temperature.

**Justification for our answer**

With above assumptions true, the only factor deciding time required to increase the temperature is Cp (Specific Heat of Water).

Heating time required:

= (Heat required for 10^{0}C temperature rise) / (Rate of heat addition)

= (m x Cp x ΔT) / (Rate of heat addition)

Where:

- Rate of heat addition is same for both containers
- m = Mass of water (Same for both containers)
- ΔT = Temperature difference between initial and final temperature (10
^{0}C for both containers) - Cp = Specific heat of water

As all the parameters are same, we need to check variation of Cp with respect to temperature.

Cp variation with temperature is non-linear. If we see the variation of Cp from 5^{0}C to 90^{0}C, Cp is decreasing with temperature with minimum value at 37^{0}C and then increasing. Following plot shows that variation.

Lets look at Cp variation between 10^{0}C to 20^{0}C (for container A) and 20^{0}C to 30^{0}C (for container B).

Above plot shows that Cp values for Container A temperature range (10^{0}C to 20^{0}C) is more that Cp values for Container B temperature range (20^{0}C to 30^{0}C). In order to estimate the actual heating time, we need to consider variation of Cp with respect to temperature (and hence with respect to time). So the calculations will be unsteady calculations. Even if we consider constant values (average of minimum and maximum), the average value Cp for container A is more than average value Cp for container B.

All above property variations are taken from **National Institute of Standards and Technology**. If you want to check property variation **Click here**. A text file containing property variations of water with respect to temperature is attached with this article. You can download that after logging with LearnCAx account.

**Let’s do a sample calculation**

Assume following values:

Heat addition rate = 3500 J/s (A typical heat addition rate of gas stove)

m = 1000 grams (Approximate weight of 1 litre water)

ΔT = 10^{0}C (Temperature difference for both containers)

Let’s calculate average value of Cp for Container A and Container B by using following chart:

Average Cp value for Container A = (4.1952 + 4.1841) / 2 = 4.1896 J/gK

Average Cp value for Container B = (4.1841 + 4.1798)) / 2 = 4.1819 J/gK

Heating time for container can be calculated using following equation:

Heating time required:

= (Heat required for 10^{0}C temperature rise) / (Rate of heat addition)

= (m x Cp x ΔT) / (Rate of heat addition)

Heating time for container A:

= (1000 x 4.1896 x 10) / (3500)

= 11.97 Seconds

Heating time for container B:

= (1000 x 4.1819 x 10) / (3500)

= 11.95 Seconds

So heating time for Container A is more than that of Container B (although this might not be signification for you ☺).

**Important Note**: Above results are valid only with the assumptions made. You may see some different results in reality. In reality when you heat water, the container is not perfectly insulated. So the heat loss rate will increase with increase in temperature. Viscosity and conductivity of water decrease with temperature and will have some impact on heating time.

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##### The Author

## Vijay Mali |
Vijay is COO and Co-Founder of CCTech & LearnCAx. Vijay's major contribution in professional career is growing CCTech from team of two people to group of 30 technologists and now CCTech is a preferred partner to many engineering industries. At CCTech, Vijay look after business development for CFD division and a member of technical review committee. Since beginning of Vijay's professional career, he has passion for education. At CCTech, he conceptualized a unique training program on CFD which was then taken by more than 500 students and most of them are now working as CFD analyst in industry. This program is now considered as benchmark for classroom training. The same passion caused birth of LearnCAx, an online education brand of CCTech. LearnCAx is first MOOC platform dedicated for CFD education. Prior to CCTech, Vijay worked with ANSYS India (formerly Fluent India) in FloWizard development team. Vijay hold M.Tech. in Aerospace Engineering from IIT Bombay. |

## Comments

2Justification:

Q equals M*C*(det T). In both the cases the mass, increase in temperature and the operating conditions are same. C increases as temperature increases which means that more heat is required to raise the temperature of a unit mass of the water as temperature increases. Since the change in C is very minimal, we can expect the time difference also to to be small.

Note : This logic will hold good only for a temperature range of 20 deg C to 40 deg C because after 40 deg C, the value of C starts to decrease as temperature decreases

2