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

Correct answer: (C) 80^° C. Heat loss = Heat gain =mS Δ θ So, m_A S_A Δ θ_A=m_B S_B Δ θ_B ⇒ 100 × S_A ×(100-90)=50 × S_B ×(90-75) 2 ~S_A=1.5 ~S_B ⇒ S_A=3/4 ~S_B Now, 100 × S_A ×(100-θ)=50 × S_B ×(θ-50) 2 × (3/4 ) ×(100-θ)=(θ-50) 300-3 θ=2 θ-100 400=5 θ ⇒ θ=80^° C

Q64 / 90 · JEE Main 2019 (11 Jan Shift 2) · Physics

When $100 \mathrm{~g}$ of a liquid $\mathrm{A}$ at $100^{\circ} \mathrm{C}$ is added to $50 \mathrm{~g}$ of a liquid $B$ at temperature $75^{\circ} \mathrm{C}$, the temperature of the mixture becomes $90^{\circ} \mathrm{C}$. The temperature of the mixture, if $100 \mathrm{~g}$ of liquid $A$ at $100^{\circ} \mathrm{C}$ is added to $50 \mathrm{~g}$ of liquid $\mathrm{B}$ at $50^{\circ} \mathrm{C}$, will be:

A
$85^{\circ} \mathrm{C}$
B
$60^{\circ} \mathrm{C}$
Correct answer
C
$80^{\circ} \mathrm{C}$
D
$70^{\circ} \mathrm{C}$

Solution

Heat loss $=$ Heat gain $=\mathrm{mS} \Delta \theta$ $\mathrm{So}, \mathrm{m}_{\mathrm{A}} \mathrm{S}_{\mathrm{A}} \Delta \theta_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{S}_{\mathrm{B}} \Delta \theta_{\mathrm{B}}$ $\begin{array}{l} \Rightarrow 100 \times \mathrm{S}_{\mathrm{A}} \times(100-90)=50 \times \mathrm{S}_{\mathrm{B}} \times(90-75) \\ 2 \mathrm{~S}_{\mathrm{A}}=1.5 \mathrm{~S}_{\mathrm{B}} \Rightarrow \mathrm{S}_{\mathrm{A}}=\frac{3}{4} \mathrm{~S}_{\mathrm{B}} \end{array}$ Now, $100 \times \mathrm{S}_{\mathrm{A}} \times(100-\theta)=50 \times \mathrm{S}_{\mathrm{B}} \times(\theta-50)$ $2 \times\left(\frac{3}{4}\right) \times(100-\theta)=(\theta-50)$ $\begin{array}{l} 300-3 \theta=2 \theta-100 \\ 400=5 \theta \Rightarrow \theta=80^{\circ} \mathrm{C} \end{array}$

Hence, the correct answer is option: 'C'.

Solution

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