Damien5232idn
Answered

IDNLearn.com provides a comprehensive solution for all your question and answer needs. Get the information you need from our community of experts who provide accurate and thorough answers to all your questions.

PLEASE HELP!
A piece of metal is heated to a temperature of 50.0°C and then placed in a calorimeter containing 50.0 g of water at 27.0°C. The water temperature increases to 35.0°C. How many joules of heat were transferred from the metal to the water? (Cwater = 4.18J/g•°C)

Sagot :

Answer: 1672 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed by water = ?  

c = heat capacity of water = [tex]4.18 J/g^0C[/tex]

Initial temperature of the water = [tex]T_i[/tex] = [tex]27^0C[/tex]

Final temperature of the water  = [tex]T_f[/tex]  = [tex]35^0C[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i=(35.0-27.0)^0C=8.0^0C[/tex]

Putting in the values, we get:

[tex]Q=50.0g\times 4.18J/g^0C\times 8.0^0C=1672J[/tex]

As heat absorbed by water is equal to the heat released by metal, the Joules of heat transferred are 1672 Joules

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.