# Question 5efd7

Mar 9, 2015

The final temperature will be $\text{60.0"^@"C}$.

The key to this problem is to realize that the heat gained by the metal will be equal to the heat lost by the water. You need the specific heats of cobalt and water, which you'll find listed as

c_("cobalt") = "0.42 J/g C"

c_("water") = "4.185 J/g C"

So, you know that

${q}_{\text{cobalt") = q_("water}}$

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

"29.4 g" * 0.42"J"/("g" * "C") * (T_f - 15.9^@"C") = "94.3 g" * 4.185"J"/("g" * "C") * (61.4^@"C" - T_f)#

Solving for ${T}_{f}$, which is the final temperature of the cobalt + water system, you'll get

$12.348 \cdot {T}_{f} - 196.33 = 24231.5 - 394.65 \cdot {T}_{f}$

$406.998 \cdot {T}_{f} = 24427.8 \implies {T}_{f} = {60.02}^{\circ} \text{C}$

Rounded to three sig figs, the answer will be

${T}_{f} = {\text{60.0}}^{\circ} C$

SIDE NOTE Notice how little the temperature of the water drops with respect to how much the temperature of the cobalt increases; this difference can be attributed to the significant difference betweeen their respective specific heats.

You need about 10 times more energy to increase the temeperature of 1 gram water by 1 degree Celsius than to increase the temperature of 1 gram of cobalt by 1 degree Celsius.