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5g of water at ${30^o}C$ and $5g$ of ice at $ - {20^o}C$ are mixed in a calorimeter. Find the final temperature, water equivalent of the calorimeter is negligible. Specific heat of ice is $0.5ca{\lg ^{ - 1o}}{C^{ - 1}}$ and latent heat of ice is $80cal$ per gram.

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Last updated date: 13th Jun 2024
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Answer
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Hint: In order to solve this question, one should be aware of the concept that there would be change in energy of ice when it melts. After mixing the ice with water it would take energy from water in order to melt. Then the final temperature of the mixture will be calculated after determining the total transfer of energy. The energy transferred by water to the ice is equal to the energy received by ice in order to melt down.

Complete step by step solution:
As we know that, the Specific heat of water, denoted by ${s_1} = 4.18J/({g^o}C) = 1cal/({g^o}C)$
As per the condition conditions given in the question we have,
Mass of water, ${m_1} = 5g$
Mass of ice, ${m_2} = 5g$
Temperature of water, ${T_1} = {30^o}C$
Temperature of ice, ${T_1} = - {20^o}C$
Latent heat of water $L = 80ca{\lg ^{ - 1}}$
Specific heat of ice, ${s_2} = 0.5ca{\lg ^{ - 1o}}{C^{ - 1}}$
Let the final temperature of the mixture be, $T$

Here in this question, the heat lost by $5g$ of water = Heat energy needed to change the temperature of ice from $ - {20^o}C$ to ${0^0}C$ + Latent heat needed to change ice at into water at ${0^0}C$ + heat absorbed by water (melted ice).
Using the formula,
${m_1}{s_1}({T_1} - T) = {m_2}{s_2}(0 - {T_2}) + {m_2}L + {m_2}{s_2}(T - {T_2})$
On putting the values we get,
$5 \times 1 \times (30 - T) = 5 \times 0.5(0 - ( - 20)) + 5 \times 80 + 5 \times 1 \times (T - (0))$
On simplifying we have,
$150 - 5T = 50 + 400 + 5T$
Taking $T$ on one side and all other factors on the other, we have,
$10T = 150 - 50 - 400$
On solving we get,
$T = - {30^o}C$

So, the final temperature of the mixture is $ - {30^o}C$.

Note: Latent Heat of Fusion is the heat per unit mass required for ice to change its phase and turn into liquid. In the question given above the Latent Heat of Fusion for the whole ice would be equated to the change in energy of the water after the ice is added to the water.