Question

How much heat is required to change 1.0 kg of ice, originally at -20.0 degrees celcius, into steam at 110.0 degrees celcius? Assume 1.0 atm of pressure

Answer 1

Answer:

$Q = 3087460 J$

Explanation:

As we know that first ice will convert into water

then water will heat to 100 degree celcius and then it will converted into vapor and then its temperature is raised to 110 degree C

First we will find the heat to take the ice at 0 degree and then convert it into water

$Q_1 = ms_{ice}\Delta T + mL$

$Q_1 = 1(2100)(0 - (-20)) + 1(335000)$

$Q_1 = 42000 + 335000 = 377000 J$

Now heat required to take it to 100 Degree then convert into vapor and then raise temperature to 110 degree

$Q_2 = 1(4186)(100 - 0) + (1)(2250000) + 1(4186)(110 - 100)$

$Q_2 = 2710460 J$

So total heat required is given as

$Q = Q_1 + Q_2$

$Q = 377000 + 2710460$

$Q = 3087460 J$