# Confusion about whether to use the specific heat of water or ice

• bluesteels
In summary: In the first question, there is no need to heat the ice as it is already at 0ºC. Therefore, the 2100 term is not needed.
bluesteels
Homework Statement
A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255°C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings?"
Relevant Equations
Q=mc delta T
Q=mL (latent heat)
My thought process of how i do the ice melting part: (note I just ignore the copper/lead part cause I already know how to do that part)

Q_ice + Q_melt + Q_liquid so, it 0.018(2100)T+0.16(4190)T+0.018(334*10^3)

but on chegg they didn't use 2100 but they just use 4190 instead and I am confused on why they did that.

Like i know it because ice melting into water but if that the case how come this problem below they don't even use it

"In a container of negligible mass, 0.200 kg of ice at an initial temperature of -40.0°C is mixed with a mass m of water that has an initial temperature of 80.0°C. No heat is lost to the surroundings. If the final temperature of the system is 28.0°C, what is the mass m of the water that was initially at 80.0°C?"

For this problem they did

0.2(2100)(40)+0.2(334*10^3)+0.2(4190)(80). if that was the case earlier then wouldn't you sub. 2100 for 4190 for this problem.

Ice melts at 0 degrees Celsius.

What is the initial temperature of the system ? (reasonably obvious : it's "in thermal equilibrium")

Assuming that the final temperature is high enough for all the ice to melt, the amount of heat (in Joules) that was transferred to the part that was initially ice was $$Q_I=0.18(0-0)+0.18(334000)+0.18(4190)(T-0)$$And the amount of heat that was transferred to the part that was initially water was $$Q_W=0.16(4100)(T-0)$$where 0C was the initial temperature of the water-ice bath.

bluesteels said:
My thought process of how i do the ice melting part: (note I just ignore the copper/lead part cause I already know how to do that part)

Q_ice + Q_melt + Q_liquid so, it 0.018(2100)T+0.16(4190)T+0.018(334*10^3)

but on chegg they didn't use 2100 but they just use 4190 instead and I am confused on why they did that.

Like i know it because ice melting into water but if that the case how come this problem below they don't even use it
For this problem, you are not heating-up any ice. You are:
- melting 0.0180kg of ice at 0ºC so it becomes water;
- then heating 0.160kg and 0.0180kg of water from 0ºC to TºC.

Chegg is correct

bluesteels said:
For this [second] problem they did

0.2(2100)(40)+0.2(334*10^3)+0.2(4190)(80). if that was the case earlier then wouldn't you sub. 2100 for 4190 for this problem.
In the second question, part of the process is to heat 0.20kg of ice from -40ºC to 0ºC (before it can melt). That’s why the answer contains “0.2(2100)(40) “.

bluesteels

## 1. What is the specific heat of water and ice?

The specific heat of water is 4.186 joules per gram per degree Celsius (J/g°C), while the specific heat of ice is 2.108 J/g°C.

## 2. Why is there confusion about which specific heat to use?

The confusion arises because the specific heat of water and ice are different, and it depends on the state of the substance (solid, liquid, or gas) and the temperature.

## 3. When should I use the specific heat of water?

You should use the specific heat of water when calculating the energy required to heat liquid water from one temperature to another.

## 4. When should I use the specific heat of ice?

You should use the specific heat of ice when calculating the energy required to cool solid ice from one temperature to another.

## 5. Can I use the same specific heat for both water and ice?

No, you cannot use the same specific heat for both water and ice because they have different values. It is important to use the correct specific heat for the substance and state you are working with to ensure accurate calculations.

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