Question

When a 190 g piece of iron at 160 degrees Celsius is placed in a 130 g aluminum calorimeter cup containing 260 g of an unknown liquid at 10 degrees Celsius, the final temperature is observed to be 38 degrees Celsius. What is the specific heat of the unknown liquid in calories/gram-degree-C?

Answer 1

Answer:

0.44 cal/(g C)

Explanation:

I assume the calorimeter is adiabatic.

First I need the specific heat capacity of iron, this is:

Cp(Fe) = 0.11 cal/(g * C)

Both bodies in the calorimeter will exchange heat until they reach equilibrium at the final temperature. The heat exchanged by each body is:

Q = m * Cp * (tfin - ti)The iron will be giving off heat (negative heat) and the liquid receiving it (positive heat, so:

-m(Fe) * Cp(Fe) *(tfin - ti(Fe)) = m(liq) * Cp(liq) * (tfin - ti(liq))

Rearranging:

$Cp(liq) = \frac{m(Fe) * Cp(Fe) *(tfin - ti(Fe))}{m(liq) * (tfin - ti(liq))}$

So:

$Cp(liq) = \frac{-190 * 0.11 *(38 - 190)}{260 * (38 - 10)} = 0.44 \frac{cal}{g C}$