# Question 23a55

##### 1 Answer
Sep 22, 2015

$q = - 2.52 \cdot {10}^{3} \text{kcal}$

#### Explanation:

The idea here is that you need to take into account the heat given off when that much water goes from steam at ${100}^{\circ} \text{C}$ to liquid at ${100}^{\circ} \text{C}$, and the heat given off when the liquid water is cooled from ${100}^{\circ} \text{C}$ to ${10}^{\circ} \text{C}$.

You need to know the enthalpy of condensation of water, which tells you how much heat is given off when $\text{1 g}$ of liquid water $100 \text{^@"C}$ is converted to steam at $100 \text{^@"C}$

$\Delta {H}_{\text{con" = -"2260 J/g}}$

Like wise, you need to know the specific heat of liquid water, which will tell you how much heat is needed to increase the temperature of $\text{1 g}$ of water by $1 \text{^@"C}$

c_"water" = 4.184"J"/("g" ^@"C") = 1"cal"/("g"^@"C")

So, start by calculating how much heat is given off when you turn that much water from steam at $100 \text{^@"C}$ to liquid at $100 \text{^@"C}$.

${q}_{1} = m \cdot \Delta {H}_{\text{con}}$

q_1 = 4.00color(red)(cancel(color(black)("kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * (-2260"J"/color(red)(cancel(color(black)("g")))) = -9.04 * 10^6"J"

Now calculate how much heat is given off when the water cools

${q}_{2} = m \cdot {c}_{\text{water}} \cdot \Delta T$

q_2 = 4.00 * 10^3color(red)(cancel(color(black)("g"))) * 1"cal"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (10 - 100)color(red)(cancel(color(black)(""^@"C"))) = -3.6 * 10^5"cal"

Convert the first heat from Joules to calories

-9.04 * 10^6color(red)(cancel(color(black)("J"))) * "1 cal"/(4.184color(red)(cancel(color(black)("J")))) = -2.16 * 10^6"cal"

The total amount of heat given off will be

${q}_{\text{total}} = {q}_{1} + {q}_{2}$

${q}_{\text{total" = -2.16 * 10""^6"cal" - 3.6 * 10""^5"cal" = -2.52 * 10""^6"cal}}$

Finally, convert the result from calories to kilocalories

-2.52 * 10^6color(red)(cancel(color(black)("cal"))) * "1 kcal"/(10^3color(red)(cancel(color(black)("cal")))) = color(green)(-2.52 * 10""^3"kcal")#