# How much heat in kcal must be added to 0.79 kg of water at room temperature (20^oC) to raise its...

## Question:

How much heat in kcal must be added to 0.79 kg of water at room temperature {eq}(20^oC) \text{ to raise its temperature to } 45^oC? {/eq}

## Heat Transfer:

The heat transferred from one substance to another can be acquired from the change of heat of one substance, along with its mass and specific heat. We determine the heat through the equation, {eq}\displaystyle q = mc\Delta T {/eq}, where *m* is the mass, *c* is the specific heat, and {eq}\displaystyle \Delta {/eq}*T* is the change in temperature.

## Answer and Explanation:

Determine the total heat, *q*, that is required to heat the water by applying the equation, {eq}\displaystyle q = mc\Delta T {/eq}. We have the mass of the water sample as {eq}\displaystyle m = 0.79\ kg {/eq}, the specific heat of {eq}\displaystyle c = 1\ kcal/kg^\circ C {/eq}, and a change in temperature of {eq}\displaystyle \Delta T = 45^\circ C - 20^\circ C = 25^\circ C {/eq}. We plug in the given values to determine the answer.

{eq}\begin{align} \displaystyle q &= mc\Delta T\\ &= (0.79\ kg)(1\ kcal/kg^\circ C)(25^\circ C)\\ &\approx\boxed{\rm 20\ kcal} \end{align} {/eq}

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from High School Physics: Help and Review

Chapter 17 / Lesson 12