At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

A 13.9 - g piece of metal ( specific heat capacity is 0.449 /g^ C)who whose temperature is 54.2 degrees * C was added to a sample of water at 13.4 degrees * C in a constant - pressure calorimeter of negligible heat capacity . If the final temperature of the water is 15.6 °C, calculate the mass of the water in the calorimeter .​

Sagot :

Answer:

26.2g = Mass of water in the calorimeter

Explanation:

The heat absorbed for the water is equal to the heat released for the metal. Based on the equation:

Q = m*C*ΔT

Where Q is heat, m is the mass of the sample, C is specific heat of the material and ΔT is change in temperature

Replacing we can write:

[tex]m_{metal}*C_{metal}*dT_{metal}=m_{water}*C_{water}*dT_{water}[/tex]

13.9g * 0.449J/g°C * (54.2°C-15.6°C) = m(H₂O) * 4.184J/g°C * (15.6°C-13.4°C)

240.9J = m(H₂O) * 9.2J/g

26.2g = Mass of water in the calorimeter

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.