Thermometry Calorimetry And Thermal Expansion Question 21
Question: 540 g of ice at 0°C is mixed with 540 g of water at 80°C. The final temperature of the mixture is [AFMC 1994]
Options:
A) 0°C
B) 40°C
C) 80°C
D) Less than 0°C
Show Answer
Answer:
Correct Answer: A
Solution:
Heat taken by ice to melt at 0°C is $ Q_{1}=mL=540\times 80=43200,cal $ Heat given by water to cool upto 0°C is $ Q_{2}=ms\Delta \theta =540\times 1\times (80-0)=43200,cal $ Hence heat given by water is just sufficient to melt the whole ice and final temperature of mixture is 0°C. Short trick : For these type of frequently asked questions you can remember the following formula $ {\theta_{\text{mix}}}=\frac{m_{W}{\theta_{W}}-\frac{m_{i}L_{i}}{c_{W}}}{m_{i}+m_{W}} $ (See theory for more details) If $ m_{W}=m_{i} $ then $ {\theta_{mix}}=\frac{{\theta_{W}}-\frac{L_{i}}{c_{W}}}{2}=\frac{80-\frac{80}{1}}{2}=0{}^\circ C $
BETA
