# Calculate the heat released for 5.00 grams, if H_2O cooling from 99 degrees Celsius to 22 degrees...

## Question:

Calculate the heat released for 5.00 grams, if {eq}\rm H_2O {/eq} cooling from {eq}99 ^\circ {/eq} Celsius to {eq}22 ^\circ {/eq} Celsius in joules and calories.

## Heat Transfer:

The heat gained by a substance translates to a change in temperature. These quantities are related through the equation, {eq}\displaystyle q = mc\Delta T {/eq}, wherein {eq}\displaystyle q {/eq} is the heat absorbed, {eq}\displaystyle m {/eq} is the mass of the object, {eq}\displaystyle c {/eq} is its specific heat, and {eq}\displaystyle \Delta T {/eq} is its change in temperature.

Determine the released heat, q, from the change in temperature of water using the equation, {eq}\displaystyle q = mc\Delta T {/eq}, wherein the mass of the sample of water is m = 5.00 g, the specific heat is c = 1 cal/g{eq}\displaystyle ^\circ {/eq}C, and the change in temperature is {eq}\displaystyle \Delta {/eq}T = 22{eq}\displaystyle ^\circ {/eq} - 99{eq}\displaystyle ^\circ {/eq} = -77{eq}\displaystyle ^\circ {/eq}C. We proceed with the solution.

{eq}\begin{align} \displaystyle q &= mc\Delta T\\ &= 5.00\ g \times 1\ cal/g^\circ C\times - 77^\circ C\\ &= \boxed{\rm - 385\ cal} \end{align} {/eq}

We convert the unit and express it in terms of joules knowing that 1 cal is 4.186 J.

{eq}\begin{align} \displaystyle q &= -385\ cal\times 4.186\ J/cal\\ &\approx \boxed{\rm -1612\ J} \end{align} {/eq}