Answered

Get the most out of your questions with the extensive resources available on IDNLearn.com. Our platform is designed to provide trustworthy and thorough answers to any questions you may have.

Energy in the amount of 420 J is added to a 35 g sample of water at a temperature of 10°C. What is the final temperature of the water?

Sagot :

Lanuelidn

The final temperature of the water, T2 = 38.57°C

Temperature can be defined as a measure of the degree of hotness or coldness of a physical object (body). Thus, it is measured with a thermometer and its units are degree Celsius (°C), Fahrenheit (°F) and Kelvin (°K).

A calorie refers to the amount of heat required to raise the temperature of a gram of water by one (1) degree Celsius (1°C).

Given the following data:

  • Quantity of energy = 420J
  • Mass = 35 grams
  • Initial temperature, T1 = 10°C

The specific heat capacity of water is 4.2 J/g°C.

To find the final temperature of the water (T2):

Mathematically, the quantity of energy (heat capacity) is given by the formula;

[tex]Q = mcdt[/tex]

Where;

  • Q represents the heat capacity or quantity of heat.
  • M represents the mass of an object.
  • C represents the specific heat capacity of water.
  • dt represents the change in temperature.

Substituting the values into the formula, we have;

[tex]420 = 3.5 \; * \; 4.2 \; * \; dt[/tex]

[tex]420 = 14.7 \; * \; dt\\\\dt = \frac{420}{14.7}[/tex]

Change in temperature, dt = 28.57°C

Next, we would solve for the final temperature by using this formula;

[tex]dt = T2 - T1[/tex]

[tex]28.57 = T_{2} - 10\\\\T_{2} = 28.57 \; + \; 10\\\\T_{2} = 38.57[/tex]

Final temperature, T2 = 38.57°C

Therefore, the final temperature of the water, T2 is equal to 38.57°C

For more information visit: https://brainly.com/question/22736508

Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. IDNLearn.com provides the best answers to your questions. Thank you for visiting, and come back soon for more helpful information.