Question

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 1

Answer : The heat change of the cold water in Joules is, $1.6\times 10^3J$

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.

$Density=\frac{Mass}{Volume}$

$Mass=Density\times Volume=1g/mL\times 45mL=45g$

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

Formula used :

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = $4.184J/g^oC$

$T_1$ = initial temperature of cold water = $24.7^oC$

$T_2$ = final temperature  = $33.4^oC$

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

$Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC$

$Q=1638.036J=1.6\times 10^3J$

Therefore, the heat change of cold water is $1.6\times 10^3J$