# Thermodynamics heat absorption problem

## Homework Statement

How much heat is absorbed in changing 2-kg of ice at -5°C to steam at 110.0°C? The Specific heats of ice, liquid water, and steam are, respectively, 2060 J/Kg x K, 4180 J/Kg x K, and 2020 J/Kg x K. The heat fusion of ice is 3.34 x 10^5 J/Kg. The heat of vaporization of water is 2.26 x 10^6 J/kg

## Homework Equations

Q=mHf
Q=mHv
Q=mC(change in temp)

## The Attempt at a Solution

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Homework Helper
You have all the information and formulas you need. Consider that you have five stages that the water has to go through in this process:

the ice must warm up from -5º C to 0º C ;
the ice must then melt into (liquid) water at a constant 0º C ;
the meltwater must now warm from 0º C to 100º C ;
the water must next vaporize into steam at a constant 100º C ; and, finally,
the steam must heat up from 100º C to 110º C .

So you will have five heat terms to compute, using the given information (they were nice to you and didn't even make you do unit conversions). You will get a total heat input in Joules.