Solveeit Logo
Login

Question

Question: 5g of water at \({30^o}C\) and \(5g\) of ice at \( - {20^o}C\) are mixed in a calorimeter. Find the ...

5g of water at 30oC{30^o}C and 5g5g of ice at −20oC - {20^o}C are mixed in a calorimeter. Find the final temperature, water equivalent of the calorimeter is negligible. Specific heat of ice is 0.5calg⁡−1oC−10.5ca{\lg ^{ - 1o}}{C^{ - 1}} and latent heat of ice is 80cal80cal per gram.

Explanation

Solution

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 s1=4.18J/(goC)=1cal/(goC){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, m1=5g{m_1} = 5g
Mass of ice, m2=5g{m_2} = 5g
Temperature of water, T1=30oC{T_1} = {30^o}C
Temperature of ice, T1=−20oC{T_1} = - {20^o}C
Latent heat of water L=80calg⁡−1L = 80ca{\lg ^{ - 1}}
Specific heat of ice, s2=0.5calg⁡−1oC−1{s_2} = 0.5ca{\lg ^{ - 1o}}{C^{ - 1}}
Let the final temperature of the mixture be, TT

Here in this question, the heat lost by 5g5g of water = Heat energy needed to change the temperature of ice from −20oC - {20^o}C to 00C{0^0}C + Latent heat needed to change ice at into water at 00C{0^0}C + heat absorbed by water (melted ice).
Using the formula,
m1s1(T1−T)=m2s2(0−T2)+m2L+m2s2(T−T2){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,
5×1×(30−T)=5×0.5(0−(−20))+5×80+5×1×(T−(0))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+5T150 - 5T = 50 + 400 + 5T
Taking TT on one side and all other factors on the other, we have,
10T=150−50−40010T = 150 - 50 - 400
On solving we get,
T=−30oCT = - {30^o}C

So, the final temperature of the mixture is −30oC - {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.