Question

6.0 g of cream at 22.9 °c are added to an insulated cup containing 150.0 g of coffee at 83.4 °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: $m_c c (T_c - T_e) = m_cre c (T_e - T_cre)$ (c cancels out)

Step2: Plug in values

$150.0(T_e - 22.9) = 6.0(83.4 - T_e)$? No: Correct: $150.0(83.4 - T_e) = 6.0(T_e - 22.9)$

Step3: Expand equation

$150.0×83.4 - 150T_e = 6T_e - 6×22.9$ → $12510 - 150T_e = 6T_e - 137.4$

Step4: Solve for $T_e$

$12510 + 137.4 = 156T_e$ → $12647.4 = 156T_e$ → $T_e = 12647.4/156 ≈ 81.1$

Answer:

81.1 °C