# If 65.34g piece of tin were to lose 754 Joules of energy in a calorimeter while experiencing a temperature change from 23.9 degrees C to 22.0 degrees C, how would you find its specific heat?

Oct 27, 2015

The values you provided are way off.

#### Explanation:

SIDE NOTE First thing first, the values you provided are very inaccurate. Using these values will produce an impossible result for the specific heat of tin, approximately $30$ times bigger than what its real value is.

WIth this being said, I will assume that the piece of tin lost $\text{75.4 J}$, instead of $\text{754 J}$. The result will not be ideal, but it will be closer to reality.

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The question wants you to determine tin's specific heat.

Before doing any calculation, try to get a clear understanding of what you need to determine.

A substance's specific heat tells you the amount of heat needed to increase the temperature of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

In your case, a mroe accurate description would sound like this - the amount of heat that must be lost by $\text{1 g}$ of a substance to decrease its temperature by ${1}^{\circ} \text{C}$.

The equation that establishes a relationship between heat lost and change in temperature looks like this

$\textcolor{b l u e}{- q = m \cdot c \cdot \Delta T} \text{ }$, where

$- q$ - heat lost by the metal
$m$ - the mass of the metal
$c$ - its specific heat
$\Delta T$ - th change in temperature, defined as final temperature minus initial temperature

The trick here is to realize that when heat is being lost by the system, $q$ will have a negative value. In this case, $\text{75.4 J}$ are being lost, so $q$ will be

$q = - \text{75.4 J}$

Since you have all the information you need to find the specific heat of tin, plug in your values in the above equation and solve for $c$

$c = \frac{- q}{m \cdot \Delta T}$

$c = \left(- \left(- \text{75.4 J"))/("65.34 g" * (22.0 - 23.9)^@"C") = color(green)(0.607"J"/("g" ""^@"C}\right)\right)$

The actual specific heat of tin is 0.21"J"/("g" ""^@"C"), so I recommend double-checking the values you have.