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Question: How much heat energy is released when 5.0 g of water at \(20^\circ C\) changes into ice at \(0^\circ...

How much heat energy is released when 5.0 g of water at 20∘C20^\circ C changes into ice at 0∘C0^\circ C ? Take specific heat capacity of water is 4⋅2Jg−1K−14 \cdot 2J{g^{ - 1}}{K^{ - 1}}, specific latent heat of fusion of ice is 336Jg−1336J{g^{ - 1}}.

Explanation

Solution

Whenever water is converted from the ice to water then there is addition of heat into the water and if there conversion of water into ice then there is release of energy from the water. Heat capacity is defined as the amount of heat energy supplied to a given mass to increase its temperature by unity.

Formula used: The formula of the heat released is,
⇒Q=mcΔT\Rightarrow Q = mc\Delta T
Where heat is Q the mass is m the change in temperature ΔT\Delta Tand the specific heat is c.
The formula of latent heat is given by,
⇒q=mC\Rightarrow q = mC
Where latent heat is q the mass of the substance m and specific latent heat of fusion is C.

Complete step by step solution:
It is given in the problem that 5g5g of water is at 20∘C20^\circ C and it changes into ice at 0∘C0^\circ C and we need to tell the amount of heat energy released in this conversion if the specific heat of water is 4⋅2Jg−1K−14 \cdot 2J{g^{ - 1}}{K^{ - 1}} and the specific latent heat of fusion of ice is 336Jg−1336J{g^{ - 1}}.
First of all let us calculate the latent heat,
The formula of latent heat is given by,
⇒Q1=mC\Rightarrow {Q_1} = mC
Where latent heat is q the mass of the substance m and specific latent heat of fusion is C.
The mass of the water is ⇒m=5g \Rightarrow m = 5g the specific latent heat capacity is 336Jg−1336J{g^{ - 1}} the latent heat is equal to,
⇒Q1=mC\Rightarrow {Q_1} = mC
⇒Q1=5×336\Rightarrow {Q_1} = 5 \times 336
⇒Q1=1680J\Rightarrow {Q_1} = 1680J………eq. (1)
Let’s calculate the heat energy released when water is converted into ic.
The formula of the heat released is,
⇒Q=mcΔT\Rightarrow Q = mc\Delta T
Where heat is Q the mass is m the change in temperature ΔT\Delta T and the specific heat is c.
The heat energy released when water at 20∘C20^\circ C changes to 0∘C0^\circ C also the specific heat of capacity is 4⋅2Jg−1K−14 \cdot 2J{g^{ - 1}}{K^{ - 1}}.
⇒Q2=mcΔT\Rightarrow {Q_2} = mc\Delta T
⇒Q2=5×4⋅2×(20−0)\Rightarrow {Q_2} = 5 \times 4 \cdot 2 \times \left( {20 - 0} \right)
⇒Q2=420J\Rightarrow {Q_2} = 420J………eq. (2)
The total heat energy released is equal to,
⇒Q=Q1+Q2\Rightarrow Q = {Q_1} + {Q_2}
Replacing the value of equation (1) and equation (2) in the above relation we get.
⇒Q=Q1+Q2\Rightarrow Q = {Q_1} + {Q_2}
⇒Q=1680+420\Rightarrow Q = 1680 + 420
⇒Q=2100J\Rightarrow Q = 2100J.

The total energy released from the water is equal to Q=2100JQ = 2100J.

Note: Whenever heat is added to the ice the molecules will start to vibrate and therefore it cannot maintain its state as ice and turn into water whereas when heat is taken away from the water the molecules stop vibrating and come closer to each other and therefore it changes from water into the ice.