Respuesta :
Answer : The heat change of the cold water in Joules is, [tex]1.6\times 10^3J[/tex]
Explanation :
First we have to calculate the mass of cold water.
As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]Mass=Density\times Volume=1g/mL\times 45mL=45g[/tex]
Now we have to calculate the heat change of cold water.
Formula used :
[tex]Q=m\times c\times (T_2-T_1)[/tex]
where,
Q = heat change of cold water = ?
m = mass of cold water = 45 g
c = specific heat of water = [tex]4.184J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature of cold water = [tex]24.7^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]33.4^oC[/tex]
Now put all the given value in the above formula, we get:
[tex]Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC[/tex]
[tex]Q=1638.036J=1.6\times 10^3J[/tex]
Therefore, the heat change of cold water is [tex]1.6\times 10^3J[/tex]
The heat change of the cold water in Joules is mathematically given as
Q=1.6*0^3J
What is the heat change of the cold water in Joules?
Question Parameter(s):
To calibrate your calorimeter cup, you first put 45 mL of cold water
hot water, temperature = 46.1 °C,
the temperature every thirty seconds over a 10 minute period.
Generally, the equation for the heat change is mathematically given as
Q=m* c* (T2-T1)
Where
Mass=d*v
Mass=1* 45mL
Mass=45g
Therefore
Q=45*4.184* (33.4-24.7)
Q=1638.036J
Q=1.6*0^3J
In conclusion, the heat change of cold water is
Q=1.6*0^3J
Read more about Charge
https://brainly.com/question/9156412
