An insulated beaker with negligible mass contains liquid water with a mass of 0.

Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.205-kg and a temperature of 75.5 degrees Celsius. How much ice at a temperature of -24.3 degrees Celsius must be dropped into the water so that the final temperature of the system will be 35.0 degrees Celsius?
Take the specific heat of liquid water to be 4190 J/kg x K, the specific heat of ice to be 2100J/kg x K and the heat fusion for water to be 3.34 x 10^5 J/kg. An insulated beaker with negligible mass contains liquid water with a mass of 0.205-kg and a temperature of 75.5 degrees Celsius. How much ice at a temperature of -24.3 degrees Celsius must be dropped into the water so that the final temperature of the system will be 35.0 degrees Celsius?
Take the specific heat of liquid water to be 4190 J/kg x K, the specific heat of ice to be 2100J/kg x K and the heat fusion for water to be 3.34 x 10^5 J/kg.
Take the specific heat of liquid water to be 4190 J/kg x K, the specific heat of ice to be 2100J/kg x K and the heat fusion for water to be 3.34 x 10^5 J/kg.

Explanation / Answer

mass of 0.205-kg

temperature of 75.5 degrees

ice at a temperature of -24.3 degrees

liquid water to be 4190 J/kg x K,

water to be 3.34 x 10^5 J/kg.

specific heat of ice to be 2100J/kg x K

The water cools from 75.5° to 30.0° Celsius.

That is a change of 45.5°C. This is the same scale as Kelvin so the change is 45.5 K

Multiply (0.205 kg)(45.5 K)(4190 J/(kgK)) = 39082.2 Joules lost by the water.

This same amount of energy is gained by the ice.

The final temperature of the ice is 35°C, and we have a certain mass (M) we are to find

: Take M*(35 K)(4190 J/(kgK)) is the amount of energy from freezing to 35°C.

Then M*(3.34 x 10^5 J/kg) is the energy during melting.

Then M*(19.5 K)(2100 J/(kgK)) is the energy from -24.3°C to 0°C.

Add those three expressions, set equal to 39082.2 Joules

and solve for M.got 1.047791 kg of ice.

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