Question

How many kJ are needed to heat 300.0 g of ice at -5.00°C into steam at 100.°C?

Answer 1

Answer:

Q = 63.9 kJ

Explanation:

Given that,

Mass of ice, m = 300 g

We need to find the heat needed to heat the given amount of ice at -5.00°C into steam at 100.°C

The specific heat of ice, c = 2.03 J/g °C

The formula for heat needed is given by :

$Q=mc\Delta T\\\\Q=300\times 2.03\times (100-(-5))\\\\Q=63945\ J\\\\Q=63.9\ kJ$

So, 63.9 kJ of heat is needed.