To calibrate your calorimeter cup, you first put 45 ml of cold water in the cup, and measure its temperature to be 24.7 °c. you then pour 47 ml of hot water, temperature = 46.1 °c, into the cup and measure the temperature every thirty seconds over a 10 minute period. you extrapolate this "cooling curve" back to the time of addition and find that the "final temperature" after mixing is 33.4 °c. what is the heat change of the cold water in joules? give your answer in standard notation. do not input units.

Answer : The heat change of the cold water in Joules is,

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.

Now we have to calculate the heat change of cold water.

Formula used :

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water =

= initial temperature of cold water =

= final temperature  =

Now put all the given value in the above formula, we get:

Therefore, the heat change of cold water is

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

answer///scientific method;