First slide

Questions on calorimetry without phase change

Question

A liquid of mass m and specific heat c is heated to a temperature T. Another liquid of mass (m/2) and specific heat 2c is heated to a temperature 2T. If these two liquids are
mixed, the resulting temperature of the mixture is:

Easy

A

(2/3)T
B

(8/5)T
C

(3/5)T
D

(3/2)T

Solution

Heat lost = Heat gained
$\Rightarrow mc (T_{f} - T) = (\frac{m}{2}) 2 c (2 T - T_{f}) \Rightarrow T_{f} = (\frac{3}{2}) T$

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