Multiple choice

When $100$g of a liquid A at $100^o$C is added to $50$g of a liquid B at temperature $75^o$C, the temperature of the mixture becomes $90^o$C. The temperature of the mixture, if $100$g of liquid A at $100^o$C is added to $50$g of liquid B at $50^o$C, will be?

  1. $80^o$C
  2. $60^o$C
  3. $70^o$C
  4. $85^o$C
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A Correct answer
Explanation

Using the heat balance equation m1*c1*(T1-T) = m2*c2*(T-T2). First case: 100*c1*(100-90) = 50*c2*(90-75) => 1000*c1 = 750*c2 => c1/c2 = 0.75. Second case: 100*0.75*(100-T) = 50*(T-50) => 7500 - 75T = 50T - 2500 => 125T = 10000 => T = 80.

AI explanation

The heat lost by liquid A equals the heat gained by liquid B, giving the equation 100 multiplied by the specific heat of A multiplied by 10 equals 50 multiplied by the specific heat of B multiplied by 15. Solving this shows the specific heat of B is 1.33 times that of A, so we substitute k = 1.33 into the second scenario. For the new mixture, 100 multiplied by (100 - T) = 50 multiplied by 1.33 multiplied by (T - 50), and solving for T gives a temperature of 80 degrees C.

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