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50 g ice at $0^{\circ} \mathrm{C}$ in insulator vessel, 50 g water of $100^{\circ} \mathrm{C}$ is mixed in it, then final temperature of the mixture is (neglect the heat loss)
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The correct answer is:
$10^{\circ} \mathrm{C}$
According to calorimeter principle,
Heat given $=$ Heat taken
$m_1 s_1(100-t)=m L+m_2 s_2(t-0)$
$\begin{aligned} 50 \times 1(100-t) & =50 \times 80+50 \times 1(t-0) \\ t & =10^{\circ} \mathrm{C}\end{aligned}$
Heat given $=$ Heat taken
$m_1 s_1(100-t)=m L+m_2 s_2(t-0)$
$\begin{aligned} 50 \times 1(100-t) & =50 \times 80+50 \times 1(t-0) \\ t & =10^{\circ} \mathrm{C}\end{aligned}$
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