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## You are on your way to a party when the host asks you to pick up a bag of ice. At the grocery store you grab a 6-kg bag that was kept at a t

Question

You are on your way to a party when the host asks you to pick up a bag of ice. At the grocery store you grab a 6-kg bag that was kept at a temperature of -4.4°C. When you get to the party, you find a large cooler to put the ice in. There is already 29 L (i.e., 29 kg) of water in the cooler at a temperature of 19°C. You toss the ice into the water and close the lid. The specific heat and latent heat of fusion for water are 4186 J/(kg⋅°C) and 3.34 × 105 J/kg, respectively. The specific heat of ice near its freezing point is 2000 J/(kg⋅°C).

Find the temperature, in degrees Celsius, of the water in the cooler after the party. Assume the ice maintains its temperature on the way to the party and the cooler is well insulated.

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Physics
6 months
2021-07-16T21:07:01+00:00
2021-07-16T21:07:01+00:00 1 Answers
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## Answers ( )

Answer:

The final temperature of the content in the cooler = 1.70°C

Explanation:

Given

Mass of ice = 6 kg

Specific heat capacity of ice = 2000 J/kg.°C

Latent Heat of fusion = (3.34 × 10⁵) J/kg = 334000 J/kg

Specific heat capacity of water = 4186 J/kg°C

Initial Temperature of ice = -4.4°C

Mass of water in the cooler = 29 kg

Specific heat capacity of water = 4186 J/kg°C

Initial Temperature of the water in the cooler = 19°C

Let the final temperature of the water in cooler be T.

(Total Heat gained by the ice, to go from -4.4°C to 0°C, then melt, the rise in temperature to the final temperature) = (The heat lost by the water in the cooler)

Note that: No heat is lost to the cooler as it is well insulated and the ice remains at -4.4°C on the way to the party till it enters the cooler.

Heat gained by the ice by the ice, to go from -4.4°C to 0°C, then melt, the rise in temperature to the final temperature is given by

1) (Heat required for ice to go from -4.4°C to 0°C) = mCΔT = (6)(2000)[0-(-4.4)] = 12000(4.4) = 52,800 J

2) (Heat required for the ice to melt at constant temperature) = mL = (6)(334000) = 2,004,000 J

3) (Heat required for the newly-formed water from ice to move from 0°C to the final temperature T) = mCΔT = (6)(4186)(T – 0) = (25,116T) J

Summed together, this total heat gives

52,800 + 2,004,000 + 25,116T

= (2,056,800 + 25116T) J

(The heat lost by the water in the cooler) = mCΔT

= (29)(4186)(19 – T) = (2,306,486 – 121,394T) J

(Total Heat gained by the ice, to go from -4.4°C to 0°C, then melt, the rise in temperature to the final temperature) = (The heat lost by the water in the cooler)

Equating this two together

2,056,800 + 25116T = 2,306,486 – 121,394T

25,116T + 121,394T = 2,306,486 – 2,056,800

146,510T = 249,686

T = (249,686) ÷ (146,510) = 1.70°C

Hope this Helps!!!