What is the final temperature of both substances when they reach thermal equilibrium.
A 12 5 g sample of granite initially at 82.
A 52 0 0 c.
A 12 5 g sample of granite initially at 82 0 c is immersed in 25 0 g of water initially at 22 0 c.
1 55 10 3 c c.
What is the final temperature of both substances when they reach thermal equilibrium.
For water c s 4 18 j g 0 c and for granite c s 0 790 j g 0 c.
In our example it will be equal to c 63 000 j 5 kg 3 k 4 200 j.
B 1 55 10 3 0 c.
A 13 0 g sample of granite initially at 82 0 o c is immersed into 26 0 g of water initially at 23 0 o c.
Let s say we want to cool the sample down by 3 degrees.
A 12 5 g sample of granite initially at 82 degrees celsius is immersed into 25 g of water initially at 22 degrees celsius.
Then δt 3 k.
What is the final temperature of both substances when they reach thermal equilibrium.
The answer to a 12 5 g sample of granite initially at 82 0 0c is immersed into 25 0 g of water initially at 22 0 0c.
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What is the final temperature of both substances when they reach thermal equilibrium.
What is the final temperature of both substances when they reach thermal equilibrium.
A 14 0g sample of granite initially at 77 0 c is immersed into 22 0g of water initially at 22 0 c.
For water cs 4 18 j oc g and for granite cs 0 790 j oc g.
C 15 7 0 c.
What is the final temperature of both substances when they reach thermal equilibrium.
What is the final temperature of both substances when they reach thermal equilibrium.
What is the final temperature of both substances when they reach thermal equilibrium.
For water c p 4 18 j g o c and for granite c p 0 790 j g o c.
A 12 5 gram sample of granite initially at 82 0 oc is immersed into 25 0 grams of water initially at 22 0 oc.
What is the final temperature of both substances when they reach thermal equilibrium.
A 12 5 g sample of granite initially at 82 0oc is immersed into 25 0 g of water initially at 22 0oc.
D 27 2 0 c.
A 12 5 g sample of granite initially at 82 0 0 c is immersed into 25 0 g of water initially at 22 0 0 c.
Calculate specific heat as c q mδt.
We will assume m 5 kg.
For water specific heat 4 18 j g degrees celsius and for granite specific heat 0 79 j g degrees celsius.
For water cs 4 18j g c and for granite cs 0 790j g c.
For water cs 4 18 j g 0c and for granite cs 0 790 j g 0c.
A 13 5g sample of granite initially at 80 0 c is immersed into 23 0g of water initially at 20 0 c.
Determine the mass of the sample.