Solveeit Logo
Login

Question

Physics Question on thermal properties of matter

When 100 g100 \,g of a liquid A at 100∘C100^{\circ}C is added to 50 g50\, g of a liquid BB at temperature 75∘C75^{\circ}C, the temperature of the mixture becomes 90∘C90^{\circ}C. The temperature of the mixture, if 100 g100\, g of liquid AA at 100∘C100^{\circ}C is added to 50 g50\, g of liquid BB at 50∘C50^{\circ}C, will be :

A

80∘C80^{\circ}C

B

60∘C60^{\circ}C

C

70∘C70^{\circ}C

D

85∘C85^{\circ}C

Answer

80∘C80^{\circ}C

Explanation

Solution

100×SA×[100−90]=50×SB×(90−75)100 \times S_{A} \times\left[100-90\right] =50\times S_{B} \times\left(90-75\right)
2SA=1.5SB2S_{A} = 1.5 S_{B}
SA=34SBS_{A} = \frac{3}{4}S_{B}
Now ,100×SA×[100−T]=50×SB(T−50), 100 \times S_{A} \times \left[100-T\right]=50 \times S_{B} \left(T-50\right)
2×(34)(100−T)=(T−50)2\times \left(\frac{3}{4}\right) \left(100-T\right) =\left(T-50\right)
300−3T=2T−100300 -3T = 2T - 100
400=5T400 =5T
T=80T = 80