Question

Question #23a55

Answer 1

`q = -2.52 * 10^3"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^@"C"` to liquid at `100^@"C"`, and the heat given off when the liquid water is cooled from `100^@"C"` to `10^@"C"`.

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

`DeltaH_"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 `"1 g"` of water by `1""^@"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""^@"C"` to liquid at `100""^@"C"`.

`q_1 = m * DeltaH_"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 * c_"water" * DeltaT`

`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_"total" = q_1 + q_2`

`q_"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")`