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A volume of 120ml of drink (half alcohol half water by mass) originally at a temperature of 25 degree C is cooled by adding 20gm ice at 0 degree C. If all the ice melts, find the final temperature of the drink. ( density of drink = 0.833gm/cx, specific heat of alcohol = 0.6cal/gm/degree C?
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?A volume of 120ml of drink (half alcohol half water by mass) origin...
A volume of 120 ml of drink (half alcohol + half water by mass) originally at a temperature of 25degreec is cooled by adding 20 gm of ice at 0degreec. If all the ice melts, find the final temperature of the drink. (Density of drink = 0.833 gm/cc, specific heat of alcohol = 0.6 cal/gm.degreec, specific heat of water = 1 cal/gm.degreec, specific latent heat of ice = 80 cal/gm).
The answer to the given question is 4-degree centigrade.
This question is part of UPSC exam. View all JEE courses
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?A volume of 120ml of drink (half alcohol half water by mass) origin...
Given information:
- Volume of drink = 120 ml
- Drink is half alcohol and half water by mass
- Initial temperature of the drink = 25°C
- Mass of ice added = 20 g
- Temperature of ice = 0°C
- Density of drink = 0.833 g/ml
- Specific heat of alcohol = 0.6 cal/g/°C

Approach:
To find the final temperature of the drink, we need to calculate the amount of heat transferred during the cooling process. The heat transferred can be calculated using the formula:

Q = mcΔT

where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

Calculations:

Step 1: Calculate the mass of the drink
Mass of drink = Volume of drink x Density of drink
Mass of drink = 120 ml x 0.833 g/ml
Mass of drink = 99.96 g

Step 2: Calculate the initial mass of alcohol and water in the drink
Since the drink is half alcohol and half water by mass, the initial mass of alcohol and water in the drink is equal.
Initial mass of alcohol = Initial mass of water = Mass of drink / 2
Initial mass of alcohol = 99.96 g / 2
Initial mass of alcohol = 49.98 g
Initial mass of water = 49.98 g

Step 3: Calculate the heat transferred from the drink to the ice
Heat transferred from the drink = Heat gained by the ice
Q(drink) = Q(ice)

Q(drink) = mcΔT(drink)
Q(ice) = mcΔT(ice)

Since the ice melts, the final temperature of the ice will be 0°C.

Q(drink) = Q(ice)
m(drink)c(drink)ΔT(drink) = m(ice)c(ice)ΔT(ice)
(mass of alcohol + mass of water)c(drink)ΔT(drink) = mass of ice x c(ice) x ΔT(ice)

Step 4: Substitute the values into the equation
(49.98 g + 49.98 g) x 0.6 cal/g/°C x (25°C - final temperature of the drink) = 20 g x 1 cal/g/°C x (0°C - 0°C)

Simplifying the equation:
(99.96 g) x 0.6 cal/g/°C x (25°C - final temperature of the drink) = 0

Step 5: Solve for the final temperature of the drink
99.96 x 0.6 x (25 - final temperature of the drink) = 0
59.976 x (25 - final temperature of the drink) = 0
1499.4 - 59.976 x final temperature of the drink = 0
-59.976 x final temperature of the drink = -1499.4
final temperature of the drink = -1499.4 / -59.976
final temperature of the drink ≈ 25°C

Result:
The final
Community Answer
?A volume of 120ml of drink (half alcohol half water by mass) origin...
If all ice melts then temperature is 4 degree celcius
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?A volume of 120ml of drink (half alcohol half water by mass) originally at a temperature of 25 degree C is cooled by adding 20gm ice at 0 degree C. If all the ice melts, find the final temperature of the drink. ( density of drink = 0.833gm/cx, specific heat of alcohol = 0.6cal/gm/degree C?
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?A volume of 120ml of drink (half alcohol half water by mass) originally at a temperature of 25 degree C is cooled by adding 20gm ice at 0 degree C. If all the ice melts, find the final temperature of the drink. ( density of drink = 0.833gm/cx, specific heat of alcohol = 0.6cal/gm/degree C? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about ?A volume of 120ml of drink (half alcohol half water by mass) originally at a temperature of 25 degree C is cooled by adding 20gm ice at 0 degree C. If all the ice melts, find the final temperature of the drink. ( density of drink = 0.833gm/cx, specific heat of alcohol = 0.6cal/gm/degree C? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?A volume of 120ml of drink (half alcohol half water by mass) originally at a temperature of 25 degree C is cooled by adding 20gm ice at 0 degree C. If all the ice melts, find the final temperature of the drink. ( density of drink = 0.833gm/cx, specific heat of alcohol = 0.6cal/gm/degree C?.
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