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Question: 30gm of water at \(30^\circ C\) is in a beaker. Which of the following, when added to water, will ha...

30gm of water at 30C30^\circ C is in a beaker. Which of the following, when added to water, will have the greatest cooling effect? (Specific heat of copper =0.1cal/gm0C= 0.1cal/gm-0^\circ C)
A) 100gms100 gms of water at 10C10^\circ C
B) 15gms15 gms of water at 0C0^\circ C
C) 3gms3 gms of water at 0C0^\circ C
D) 18gms18 gms of water at 0C0^\circ C

Explanation

Solution

For any question related to cooling effect, use the principle of calorimeter which states that heat energy lost by the body (hot) is mathematically equal to the heat energy gained by the body (cooled) attains the same temperature. Heat transfer occurs until both the bodies attain the same temperature.

Complete step by step solution:
For option (A), 100gms100 gms of water at 10C10^\circ C is added to 30gms30 gms of water at 30C30^\circ C.
By using the principle of calorimeter, heat energy lost by the hot body is equal to the heat energy gained by the cold body and this heat transfer occurs until both the bodies attain the same temperature.
Given that Specific heat of copper =0.1cal/gmC = 0.1cal/gm- ^\circ C
Heat transfer is given by
H=mcΔt{{H = mc\Delta t}}
Substituting the given values in above formula, we have
100×1×(T100)=30×1×(30T) T=19013=14.60C \Rightarrow {{100}} \times {{1}} \times {{(T - 100) = 30}} \times {{1}} \times {{(30 - T)}} \\\ \Rightarrow {{T = }}\dfrac{{{{190}}}}{{{{13}}}}{{ = 14}}{{.}}{{{6}}^{{0}}}{{C}}
For option (B), 15gms15 gms of water at 0C0^\circ C is added to 30gms30 gms of water at 30C30^\circ C
Heat transfer is given by
H=mcΔt{{H = mc\Delta t}}
On substituting the values in above formula, we have
15×1×(T0)=30×1×(30T) T=603=280C \Rightarrow {{15}} \times {{1}} \times {{(T - 0) = 30}} \times {{1}} \times {{(30 - T)}} \\\ \Rightarrow {{T = }}\dfrac{{{{60}}}}{3}{{ = 2}}{{{8}}^{{0}}}{{C}}
For option (C), 3gms3 gms of water at 0C0^\circ C is added to 30gms30 gms of water at 30C30^\circ C
Heat transfer is given by
H=mcΔt{{H = mc\Delta t}}
Substituting the given values in above formula, we have
3×80×(3T0)=30×1×(30T) T=22011=200C  \Rightarrow 3 \times 80 \times {{(3T - 0) = 30}} \times {{1}} \times {{(30 - T)}} \\\ \Rightarrow {{T = }}\dfrac{{{{220}}}}{{11}}{{ = 2}}{{{0}}^{{0}}}{{C}} \\\
For option (D), 18gms18 gms of water at 0C0^\circ C is added to 30gms30 gms of water at 30C30^\circ C
Heat transfer is given by
H=mcΔt{{H = mc\Delta t}}
Substituting the given values in above formula, we have
18×0.1×(T0)=30×1×(30T) T=1505.3=28.30C  \Rightarrow 18 \times 0.1 \times {{(T - 0) = 30}} \times {{1}} \times {{(30 - T)}} \\\ \Rightarrow {{T = }}\dfrac{{{{150}}}}{{5.3}}{{ = 28}}{{.}}{{{3}}^{{0}}}{{C}} \\\
Here, the resultant minimum temperature or maximum cooling effect is for 14.61C14.61^\circ C.

Therefore, option (A) is the correct choice.

Note: Explanation for principle of calorimeter: When two bodies of different temperatures are placed in physical contact with one another then the heat transfer from the body having higher temperature to the body having lower temperature until thermal equilibrium is maintained between them. It is very important to note that this principle indicates the law of conservation of energy.