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Questions based on phase change

Question

19 g of water at 30⁰C and 5 g of ice at $-$ 20⁰C are mixed together in a calorimeter. What is the final temperature of the mixture? (Take, specific heat of ice = 0.5 cal $g^{- 1}$ ⁰ $C^{- 1}$ and latent heat of fusion of ice = 80 cal $g^{- 1}$ )

Moderate

A

0⁰C
B

$-$ 5⁰C
C

5⁰C
D

10⁰C

Solution

Here, specific heat of ice, $c_{ice} = 0.5 {calg}^{- 1 o} C^{- 1}$
Specific heat of water, $c_{water} = 1 {calg}^{- 1 o} C^{- 1}$
Latent heat of fusion of ice , $L_{ice} = 80 {calg}^{- 1}$
Here, ice will absorb heat while hot water will release it.
Let T be the final temperature of the mixture.
Assuming water equivalent of calorimeter to be neglected.
Heat given by water, $Q_{1} = m_{water} c_{water} Δ T$
$= 19 \times 1 \times (30 - T) = 570 - 19 T \dots (i)$
Heat absorbed by ice,
$Q_{2} = m_{ice} \times c_{ice} \times [0 - (- 20)] + m_{ice} \times L_{f ice}$ $+ m_{ice} \times c_{water} \times (T - 0)$
$= 5 \times 0.5 \times 20 + 5 \times 80 + 5 \times 1 \times T = 450 + 5 T \dots (ii)$
According to principle of calorimetry, $Q_{1} = Q_{2}$
$570 - 19 T = 450 + 5 T \Rightarrow T = \frac{120}{24} = 5^{\circ} C$

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