# If 74.5 grams of a hot metal was added to 56.4 grams of water and the temperature of the water...

## Question:

If {eq}74.5 {/eq} grams of a hot metal was added to {eq}56.4 {/eq} grams of water and the temperature of the water rose from {eq}20.4 {/eq} to {eq}33.4 {/eq} degrees, how many {eq}\mathrm{KJ} {/eq} of energy was lost by the hot metal?

## Heat Transfer:

In an isolated system of two substances, the heat gained by a substance corresponds to the heat lost by the other. With this, we can determine the heat lost by computing how much heat is gained by the other substance, using the heat transfer equation, {eq}\displaystyle q = mc\Delta T {/eq}, where m is the mass, c is the specific heat and {eq}\displaystyle \Delta T {/eq} is the change in temperature of the substance.

Determine the energy lost by the hot metal by finding how much heat is gained by the water using the equation, {eq}\displaystyle q = mc\Delta T {/eq}, where m = 56.4 g is the mass of water, c = 4.186 J/g{eq}\displaystyle ^\circ C {/eq} is the specific heat and {eq}\displaystyle \Delta {/eq}T = 33.4 - 20.4 = 13.0{eq}\displaystyle ^\circ C {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle q&= mc\Delta T\\ &= 56.4\ g\times 4.186\ J/g^\circ C\times 13.0^\circ C\\ &\approx 3069\ J\\ &\approx\boxed{ 3.07\ kJ} \end{align} {/eq}

The hot metal lost {eq}\displaystyle 3.07\ kJ {/eq} of heat energy.