# Mass of Ice required to achieve a certain final Temperature

• CrosisBH
In summary, to make the temperature of the water 40°C, 2 kilograms of ice at a temperature of -23.9°C must be dropped into the water.
CrosisBH

## Homework Statement

An insulated beaker with negligible mass contains a mass of 0.350 kg of water at a temperature of 76.5 °C.
How many kilograms of ice at a temperature of − 23.9 ∘C must be dropped in the water to make the final temperature of the system 40.0 ∘C?
Take the specific heat for water to be 4190 J/(kg⋅K) , the specific heat for ice to be 2100 J/(kg⋅K) , and the heat of fusion for water to be 334 kJ/kg .

## Homework Equations

$$Q = mc\Delta T$$
$$Q = mL_f$$

## The Attempt at a Solution

I recognized what exactly happened which each part of the system.

Ice at T_ice is heated to 0°C -> Ice is transformed to water -> The water is heated to T_f
Water at T_water is cooled to T_f

So setting up the equation
$$Q_{ice} + Q_{ice\rightarrow water} + Q_{water(ice)} + Q_{water} = 0$$
Expanding
$$m_{ice} c_{ice} (0°C - T_{ice}) + m_{ice}L_f + m_{ice}c_{water}(T_f - 0°C)+m_{water}c_{water}(T_f-T_{water}) = 0$$
I then solved for m_ice,
$$m_{ice} = \frac{m_{water} c_{water} T_i - m_{water} c_{water} T_f}{L_f + c_{water} T_f - c_{ice} T_{ice}}$$

And then I plugged all the values into Desmos:

L_f was converted from kJ/kg to J/kg. Getting -0.170 kg when doing 3 significant figures. Obviously there's something wrong here. I went through the algebra multiple times, so I think there's something wrong with my Physics.

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CrosisBH said:
something wrong with my Physics.
No, just lack of care in notation. Do you notice there's a given number you have not used?

scottdave
haruspex said:
No, just lack of care in notation. Do you notice there's a given number you have not used?
Yep. It was bad notation. Dumb me used T_i as both temperature of ice and the initial temperature of water in Desmos. I guess this is a lesson for me to properly name my variables so no confusion can stir from me lol.

Tom.G
Well I find out something new everyday. That Desmos looks impressive. It may be overkill for this type of math problem. Perhaps that is how one of your variables got lost.

## 1. How do you calculate the mass of ice required to achieve a certain final temperature?

The formula for calculating the mass of ice required is m = (mwater * (Tfinal - Tinitial)) / (Tmelt - Tfinal), where mwater is the mass of water, Tfinal is the desired final temperature, Tinitial is the initial temperature of the water, and Tmelt is the melting point of ice.

## 2. How does the initial temperature of the water affect the mass of ice required?

The higher the initial temperature of the water, the less mass of ice is required to reach the desired final temperature. This is because the initial temperature of the water is closer to the desired final temperature, therefore less energy is needed to raise the temperature of the water.

## 3. What is the significance of the melting point of ice in this calculation?

The melting point of ice is the temperature at which ice changes from a solid to a liquid. In this calculation, it determines the amount of energy needed to melt the ice, which in turn affects the amount of energy available to raise the temperature of the water.

## 4. Is the mass of ice required affected by the type of container used?

No, the mass of ice required is not affected by the type of container used. This calculation only takes into account the mass and temperature of the water, as well as the melting point of ice.

## 5. Can this calculation be used for any desired final temperature?

Yes, this calculation can be used for any desired final temperature as long as the initial temperature of the water is lower than the desired final temperature and the final temperature is lower than the melting point of ice.

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