# When 86.7 g of water at a temperature of 73 degree C is mixed with an unknown mass of water at a...

## Question:

When 86.7 *g* of water at a temperature of {eq}73^{\circ}C {/eq} is mixed with an unknown mass of water at a temperature of {eq}22.3^{\circ}C {/eq} the final temperature of the resulting mixture is {eq}61.7^{\circ}C {/eq}. What was the mass of the second sample of water? (Specific heat of water 4.184 {eq}J g^{-1} \ ^{\circ}C^{-1} {/eq})

## Heat Transfer Equation:

The heat transferred from a material to another material can be expressed through an equation written as {eq}\displaystyle q = mc\Delta T {/eq}, where {eq}\displaystyle m {/eq} is the mass, {eq}\displaystyle c {/eq} is the specific heat, and {eq}\displaystyle \Delta T {/eq} is the change in temperature. We can see that the change in temperature is proportional to the heat transferred, which has a proportionality constant of the mass and the specific heat of the material.

## Answer and Explanation:

Determine the mass of the unknown, colder, sample, {eq}\displaystyle m_{cold} {/eq}, of water by equating the heat lost by the hotter water, {eq}\displaystyle -q_{hot} {/eq}, and the heat gained by the colder water, {eq}\displaystyle q_{cold} {/eq}, or {eq}\displaystyle -q_{hot} = q_{cold} {/eq}. We know that the heat transfer equation is given as {eq}\displaystyle q = mc\Delta T {/eq}, where {eq}\displaystyle m {/eq} is the mass, {eq}\displaystyle c {/eq} is the specific heat, and {eq}\displaystyle \Delta T {/eq} is the change in temperature. We are given the following values:

- {eq}\displaystyle m_{hot} = 86.7\ g {/eq}

- {eq}\displaystyle \Delta T_{hot} = 61.7^\circ C - 73 ^\circ C = -11.3 ^\circ C {/eq}

- {eq}\displaystyle \Delta T_{cold} = 61.7^\circ C - 22.3 ^\circ C =39.4 ^\circ C {/eq}

- {eq}\displaystyle c = 4.184\ \rm{J/g ^\circ C} {/eq}

We proceed with the solution.

{eq}\begin{align} \displaystyle -q_{hot} &= q_{cold}\\ -m_{hot}c\Delta T_{hot} &= m_{cold}c\Delta T_{cold}\\ - \frac{m_{hot}\Delta T_{hot} }{\Delta T_{cold}} &= m_{cold}\\ - \frac{86.7\ g\times -11.3 ^\circ C }{39.4 ^\circ C} &= m_{cold}\\ 24.9 &\approx m_{cold} \end{align} {/eq}

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from High School Physics: Help and Review

Chapter 17 / Lesson 12