A 30.9-g sample of water at 290. K is mixed with 50.2 g water at 330. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. Temperature = K
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
![The problem presents a scenario involving the mixing of two water samples with different masses and initial temperatures. You are tasked with calculating the final temperature of the mixture assuming no heat is lost to the surroundings.
**Problem Statement:**
A 30.9-g sample of water at 290 K is mixed with a 50.2-g sample of water at 330 K. Calculate the final temperature of the mixture, assuming no heat loss to the surroundings.
**Input Field:**
- Temperature = \_\_\_\_ K
**Interactive Features:**
- A "Submit Answer" button for entering your solution.
- A "Try Another Version" button to explore additional problem variations.
- An indication of "3 item attempts remaining" to inform you of the number of tries allowed for this question.
**Status Indicators:**
- An orange button labeled "Not Visited" signifies that this question has not yet been attempted.
Ensure to use the principle of conservation of energy to solve this problem:
\[ \text{(Mass of Sample 1)} \times \text{(Specific Heat Capacity of Water)} \times \text{(Change in Temperature of Sample 1)} + \]
\[ \text{(Mass of Sample 2)} \times \text{(Specific Heat Capacity of Water)} \times \text{(Change in Temperature of Sample 2)} = 0 \]
This approach keeps the total energy balance of the system, considering the heat gained by one sample is equal to the heat lost by the other.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa82188d-7c1b-4aa7-8add-9da07b86f459%2F1cfbca5c-a7d3-4fc4-830c-a80cfa06e20e%2Ftkgbdt_processed.png&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
