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A glass beaker contains $200 \mathrm{~g}$ of carbonated water initially at $20^{\circ} \mathrm{C}$. How much ice should be added to obtain the final temperature of $0^{\circ} \mathrm{C}$ with all ice melted, if the initial temperature of ice is $-10^{\circ} \mathrm{C}$. Neglect heat capacity of glass.
[Take, $C_{\text {water }}=4190 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$,
$$
C_{\text {ice }}=2100 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}, L_F=3.34 \times 10^5 \mathrm{~J} / \mathrm{kg} \text { ] }
$$
Options:
[Take, $C_{\text {water }}=4190 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$,
$$
C_{\text {ice }}=2100 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}, L_F=3.34 \times 10^5 \mathrm{~J} / \mathrm{kg} \text { ] }
$$
Solution:
2269 Upvotes
Verified Answer
The correct answer is:
$47 \mathrm{~g}$
Heat lost by carbonated water $=$ Heat gained by ice
Let, $m_{\hat{i}}=$ mass of ice melted and $m_w=$ initial mass of carbonated water.
$$
\begin{aligned}
& m_w c_w(20-0) & =m_i L_f+m_i\left(c_i\right)[0-(-10)] \\
\Rightarrow \quad & & 200 \times 10^{-3} \times 4190 \times 20 \\
& = & m_i\left(3.34 \times 10^5+2100 \times 10\right) \\
\Rightarrow & & m_i=47 \times 10^{-3} \mathrm{~kg} \text { or } m_i=47 \mathrm{~g}
\end{aligned}
$$
Let, $m_{\hat{i}}=$ mass of ice melted and $m_w=$ initial mass of carbonated water.
$$
\begin{aligned}
& m_w c_w(20-0) & =m_i L_f+m_i\left(c_i\right)[0-(-10)] \\
\Rightarrow \quad & & 200 \times 10^{-3} \times 4190 \times 20 \\
& = & m_i\left(3.34 \times 10^5+2100 \times 10\right) \\
\Rightarrow & & m_i=47 \times 10^{-3} \mathrm{~kg} \text { or } m_i=47 \mathrm{~g}
\end{aligned}
$$
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