Question

7.0 g of cream at 10.4 °c are added to an insulated cup containing 100.0 g of coffee at 78.7 °c. calculate the equilibrium temperature of the coffee. you may assume no heat is lost to the cup or surroundings, and that any physical properties of cream and coffee you need are the same as those of water. be sure your answer has 3 significant digits.

Explanation:

Step1: Define heat transfer equation

Heat lost by coffee = Heat gained by cream: $m_c c (T_c - T_e) = m_cre c (T_e - T_cre)$
Cancel $c$ (same for both): $m_c (T_c - T_e) = m_cre (T_e - T_cre)$

Step2: Plug in values

$m_c=100.0g, T_c=78.7^\circ C, m_cre=7.0g, T_cre=10.4^\circ C$
$100(78.7 - T_e) = 7(T_e - 10.4)$

Step3: Expand and solve for $T_e$

$7870 - 100T_e = 7T_e - 72.8$
$7870 + 72.8 = 107T_e$
$7942.8 = 107T_e$
$T_e = \frac{7942.8}{107} \approx 74.23$

Step4: Round to 3 sig figs

$T_e \approx 74.2^\circ C$

Answer:

74.2°C