Get Answers to all your Questions

header-bg qa

Question : Heat Transfer in a System

Consider a system that consists of 200 g of water initially at 25°C. The system absorbs 1500 J of heat from the surroundings. Calculate the final temperature of the water assuming no phase change occurs.

Given:
Specific heat capacity of water (C or c) =4.18J/g\degree C

Option: 1

\mathrm{ 25.004 ^{\circ}C}


Option: 2

\mathrm{20.214 }\degree C


Option: 3

\mathrm{28.696 \degree C}


Option: 4

 \mathrm{26.794 \degree C}


Answers (1)

best_answer

The change in internal energy (U) of the system can be calculated using the formula:

\mathrm{U=m c T}

Where:         m is the mass of the substance (in grams)
                     c is the specific heat capacity (in J/(g°C))
                     T is the change in temperature (in °C)

Since no phase change occurs, we can assume that the entire heat absorbed
is used to increase the temperature.

Step 1: Calculate the change in temperature (T):

\mathrm{\begin{gathered} Q=m c T \\ T=\frac{Q}{m c} \end{gathered}}

Substitute the given values:

\mathrm{\begin{gathered} T=\frac{1500 \mathrm{~J}}{(200 \mathrm{~g}) \times\left(4.18 \mathrm{~J} /\left(\mathrm{g}^{\circ} \mathrm{C}\right)\right)} \\ T \approx 1.794^{\circ} \mathrm{C} \end{gathered}}

Step 2: Calculate the final temperature: The final temperature\mathrm{T_{f}} can be calculated by adding the change in temperature (T) to the initial temperature:

\mathrm{T_f=T_i+T}

Substitute the values:

\mathrm{\begin{gathered} T_f=25^{\circ} \mathrm{C}+1.794^{\circ} \mathrm{C} \\ T_f \approx 26.794^{\circ} \mathrm{C} \end{gathered}}

The final temperature of the water is approximately 26.794 °C. article amsmath

Posted by

Gaurav

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE