# A 455-g chunk of iron is removed from an oven and plunged into 385 g water in an insulated...

## Question:

A {eq}455-g {/eq} chunk of iron is removed from an oven and plunged into {eq}385\ g {/eq} water in an insulated container. The temperature of the water increases from {eq}27 ^\circ {/eq} Celsius to {eq}89 ^\circ {/eq} Celsius. If the specific heat of iron is {eq}\rm 0.449\ J/ g \cdot ^\circ Celsius {/eq}, what must have been the original temperature of the iron?

## Heat Transfer:

The heat lost by one substance is gained by another in an isolated environment. We must take note that we can quantify the heat lost or gained by one substance using the equation, {eq}\displaystyle q = mc\Delta T {/eq}. In this equation, we have *m* as the mass. *c* as the specific heat, and {eq}\displaystyle \Delta {/eq}*T* as the change in temperature.

## Answer and Explanation:

Determine the initial temperature of the metal, {eq}\displaystyle T_{i_{iron}} {/eq}, by equating the heat lost by the iron chunk to the heat gained by the water,

{eq}\displaystyle -q_{iron} = q_{water} {/eq}

We assign a negative value to the heat of iron since it lost heat. Now, we must take note that the heat transfer is given by the equation,

{eq}\displaystyle q = mc\Delta T {/eq}

We are given the following values:

- {eq}\displaystyle m_{iron} = 455\ g {/eq}

- {eq}\displaystyle m_{water} = 385\ g {/eq}

- {eq}\displaystyle \Delta T_{water}= 89 ^\circ C - 27 ^\circ C = 62 ^\circ C {/eq}

- {eq}\displaystyle c_{iron} = 0.449\ J/ g ^\circ C {/eq}

- We must take note thatn {eq}\displaystyle c_{water} = 4.186\ J/g ^\circ C {/eq}

- {eq}\displaystyle \Delta T_{metal} = 89^\circ C - T_{i_{iron}} {/eq}

We proceed with the solution.

{eq}\begin{align} \displaystyle -q_{iron} &= q_{water}\\ -m_{iron}c_{iron}\Delta T_{iron} &= m_{water}c_{water}\Delta T_{water}\\ \Delta T_{iron} &=-\frac{ m_{water}c_{water}\Delta T_{water}}{m_{iron}c_{iron}}\\ 89 ^\circ C - T_{i_{iron}} &= - \frac{385\ g\times 4.186\ J/g ^\circ C\times 62 ^\circ C}{455\ g\times 0.449\ J/ g ^\circ C}\\ T_{i_{iron}} &= 89 ^\circ C + \frac{385\ g\times 4.186\ J/g ^\circ C\times 62 ^\circ C}{455\ g\times 0.449\ J/ g ^\circ C}\\ &\approx \boxed{\rm 578 ^\circ C} \end{align} {/eq}

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from High School Physics: Help and Review

Chapter 17 / Lesson 12