Answer to Question #88905 in General Chemistry for Hripsime

Question

15 grams of metal X are heated to 250°C. If it is placed into 30 grams of water at 15°C, what will be the final temperature of the water?

Answer 1

At thermal equilibrium

$q_{water}=q(Metal)$

$q_{water}=c_wm_w(T_f-T_w)= 4.184\times30\times (T_f-15)$ $q_{metal}=c_m\times m_m\times (T_f-T_m)$

As we are not give the metal, we can take some metal as an example and show the calculations. Let's take Al

$c_{Al}= 0.900\frac{J}{g\times K}$

Find final temperature of water

$4.184\times 30\times(T_f-15)= -15\times 0.900\times(T_f-250)$

$125.52\times T_2-1882.8=-13.5\times T_2+3375$

$139.02\times T_f= 5257.8$

$T_f= 37.8^\circ C$