# Question 0efc9

Jan 26, 2015

The final temperature will be $\text{20.7"^@"C}$.

In order to solve this problem you need the specific heat of both copper, which is listed at $\text{0.386 J/g"* "K}$, and water, which is listed at $\text{4.18 J/g" * "K}$.

So, the key to this problem is the fact that the heat lost by the hot metal when placed in the cooler water is equal to the heat gained by the water. So,

$- {q}_{\text{copper") = q_("water}}$ (the minus sign indicates heat lost).

This means that

$- {m}_{\text{copper") * c_("copper") * DeltaT_("copper") = m_("water") * c_("water") * DeltaT_("water}}$

Let's assume that the final temperature is ${\text{T}}_{f}$. The above equation becomes

"-45.5 g" * 0.386 "J"/("g" * "K") * ("T"_(f) - 99.8) = "152 g" * 4.18 "J"/("g" * "K") * ("T"_(f) - 18.5)#

$- 17.56 \cdot \left({\text{T"_(f) - 99.8) = 653.4 * ("T}}_{f} - 18.5\right)$

Solving this equation for ${\text{T}}_{f}$ will give

$13507.5 = 653 \cdot {T}_{f} \implies {T}_{f} = \frac{13507.5}{653} = {20.7}^{\circ} \text{C}$

Notice how extreme the drop in temperature is for the metal and how small the increase in temperature is for the water; this happens because of the bigger specific heat value water has.