# A 30.5 g sample of an alloy at 94.8 degrees C is placed into 52.1 g water at 22.8 degrees C in an...

## Question:

A {eq}30.5\ g {/eq} sample of an alloy at {eq}94.8 ^\circ C {/eq} is placed into {eq}52.1\ g {/eq} water at {eq}22.8 ^\circ C {/eq} in an insulated coffee cup. The heat capacity of the coffee cup (without the water) is {eq}\rm 9.2\ J/K {/eq}. If the final temperature of the system is {eq}31.1 ^\circ C {/eq}, what is the specific heat capacity of the alloy? ({eq}C {/eq} of water is {eq}\rm 4.184\ J/g \cdot K {/eq})

## Heat Transfer:

In an insulated environment, the heat transfer between two substances only considers the heat exchange between them. Therefore, we can equate the heat lost by one substance to the heat gained by the other, such that {eq}\displaystyle q_{lost}= q_{gained} {/eq}.

## Answer and Explanation:

Determine the heat capacity of the alloy, {eq}\displaystyle c_{alloy} {/eq}, by equating the heat lost by the metal to the heat gained by the water, {eq}\displaystyle -q_{alloy} = q_{water} {/eq}. We assign a negative sign for a heat loss by a substance. We must take note that the heat transfer equation is given by {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. We are given the following values:

- {eq}\displaystyle m_{alloy} = 30.5\ g {/eq}

- {eq}\displaystyle \Delta T_{alloy} = 31.1 ^\circ C- 94.8 ^\circ C = -63.7 ^\circ C {/eq}

- {eq}\displaystyle m_{water} = 52.1\ g {/eq}

- {eq}\displaystyle \Delta T_{water} = 31.1 ^\circ C - 22.8^\circ C = 8.3 ^\circ C {/eq}

We must take note that the heat capacity of water is {eq}\displaystyle c_{water} = 1\ cal/g ^\circ C {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle -q_{alloy} &= q_{water}\\ -m_{alloy} c_{alloy} \Delta T_{alloy} &= m_{water}c_{water}\Delta T_{water}\\ c_{alloy} &= -\frac{m_{water}c_{water}\Delta T_{water}}{m_{alloy}\Delta T_{alloy}}\\ &= -\frac{52.1\ g\times 1\ cal/g ^\circ C\times 8.3 ^\circ C}{30.5 \ g\times -63.7 ^\circ C}\\ &\approx\boxed{0.223\ cal/g ^\circ C} \end{align} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from High School Physics: Help and Review

Chapter 17 / Lesson 12