# Question #b5482

Apr 3, 2017

The heat requirement is 26.72 kJ to melt the ice and 23.408 kJ to warm the water, for a total of 50.128 kJ.

#### Explanation:

There are two parts to this process. First, the ice must melt. This happens at a constant 0°C temperature. Then, the water is warmed to 70°C

For the first part, we need the heat of melting for ice. This can be looked up, and turns out to be 334 joules per gram.

So, to melt 80.0 g requires $80.0 \times 334 = 26 720 J$

Then, to heat the water will require 4.18 J per gram and per degree Celsius. This value is known as the specific heat of water, and is symbolized $c$.

In equation form:

$E = m \times c \times \Delta t$

$E = 80.0 \times 4.18 \times 70 = 23 408 J$

So, altogether, 50 128 J (50.128 kJ) to do the job.