You can use this formula to help:
Where:
C = specific heat
q = heat
m = mass
t = temperature
What we know:
C = unknown
q = 956 J
m = 45 g
change in t = 12°C because 37°C - 25°C = 12°C
Plug known values into the formula:
C = 956 J / (45 g) (12°C) and we are left with a specific heat of 1.77J/g°C
Now, convert Joules to calories and then you get:
Answer: A. 0.423 cal/g°C
Answer:
a) 1 litre of 10% solution and 7 litre of 20% solution
b) 1.67 litres of 50% solution and 8.33 litres of the 20% solution
c) 3.75 litres of 50% solution and 6.25 litres of 20% solution
Explanation:
Given:
chemist needs = 10 liters of a 25% acid solution
Concentration of three solutions that are to be mixed = 10%, 20% and 50%.
Solution:
A) Use 2 liters of the 50% solution
Let us mix this with 10% and 20% solution
They will have to equal 8 litres
Let x=20% solution
Then (8-x) =10%
So the equation becomes,
10%(8-x)+ 20%x+50%(2)=25(10)
(0.1)(8-x) +0.2x+0.50(2)= 0.25(10)
0.8-0.1x+0.2x+1.0=2.5
0.2x-0.1x=2.5-0.8-1.0
0.1x=0.7
x= 7
so, 8-x = 8 -7= 1 litre of 10% solution and 7 litre of 20% solution
B)Use as little as possible of the 50% solution
Let x be the amount of 50% solution.
Then(10-x) be the 20% solution
Now the equation becomes,
50%(x)+20%(10-x)=25%(10)
0.50x+0.2(10-x)=0.25(10)
0.5x+2.0-0.2x=2.5
0.3x=2.5-2.0
0.3x=0.5
x=1.67
now (10-x)=(10-1.67)=8.33
so there will be 1.67 litres of 50% solution and 8.33 litres of the 20% solution
c) ) Use as much as possible of the 50% solution
Let x be the amount of 50% solution.
Then(10-x) be the 20% solution
Now the equation becomes,
50%(x)+10%(10-x)=25%(10)
0.50x+0.1(10-x)=0.25(10)
0.5x+1.0-0.1x=2.5
0.4x=2.5-1.0
0.4x=1.5
x=3.75
Now, (10-x)=(10- 3.75)=6.26
So there will be 3.75 litres of 50% solution and 6.25 litres of 20% solution