Two liquids,A and B,are mixed together,and the resulting temperature is 22 $\circ$ C.If liquid A has mass m and was initially at temperature 35 $\circ$ C,and liquid B has mass 3m and was initially at temperature 11 $\circ$ C,calculate the ratio of the specific heats of A divided by B.

Question

Quizplus · Accepted Answer

The final temperature of the mixture is 22°C. Using the principle of conservation of energy, the heat lost by liquid A will be equal to the heat gained by liquid B. The formula for heat transfer is $Q = mc\Delta T$, where $Q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature.For liquid A, $Q_A = mc_A(T_f - T_i) = mc_A(22 - 35) = -13mc_A$.For liquid B, $Q_B = 3mc_B(T_f - T_i) = 3mc_B(22 - 11) = 33mc_B$.Setting $Q_A = -Q_B$, we get $-13mc_A = -33mc_B$, simplifying to $c_A/c_B = 33/13 = 2.5$.

Two Liquids,A and B,are Mixed Together,and the Resulting Temperature Is ∘\circ∘

Two Liquids,A and B,are Mixed Together,and the Resulting Temperature Is $\circ$