# How Many Grams of Ice Will Melt When Iron Meets Ice?

• Kaleb
In summary, the problem involves a 50-gram chunk of iron with a temperature of 80 degrees Celsius being dropped into ice at 0 degrees Celsius. The task is to determine the amount of ice that will melt. Using the equation Q=mc∆T, where Q is the quantity of heat, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature, the solution is found by setting the heat gained by the ice equal to the heat lost by the iron. The resulting equation is .11cal/g*C * 50g(0-80C) = .5*y(0-0C), which simplifies to 440cal = .5*y(80
Kaleb
[SOLVED] Thermo Energy. Iron + Ice

## Homework Statement

A 50-g chunk of 80 degree C iron is dropped into a cavity in a very large block of ice at 0 degrees C. How many grams of ice will melt? (The specific heat capacity of iron is .11 cal/g*C)

## Homework Equations

Q = cmdeltaT
Quantity of heat = heat capacity * mass * change in temp

also i know that it takes 80 calories to go just from ice to water

## The Attempt at a Solution

All my attempts have met with failure:

1) .11cal/g*C * 50g(x-80C)=.5*y(x-0C) (i don't know the mass of ice so I am clueless on how to solve this)

2) .11cal/g*c *50g(0-80C)=.5*y(0-0C) (assuming that the ice will equalize the final temp but it leaves the right equation = 0 which makes it pointless.

There is a re-occurring number that I keep having and its 440 cal/g. Any and all help is appreciated.

I figured it out, my second equation was correct. Thanks though!

Last edited:
Q=mc∆T.11cal/g*C *50g(0-80C) = 440cal440cal/80C = .11cal/g*C *50g440cal/50g = .88cal/g*C = mm = 440cal/50g/.88cal/g*Cm = 500g

I would approach this problem by first identifying the known values and the unknown value. The known values are the mass of the iron (50 g), its initial temperature (80 degrees C), the specific heat capacity of iron (0.11 cal/g*C), and the final temperature after the iron is dropped into the ice (0 degrees C). The unknown value is the mass of ice that will melt.

Using the formula Q = cmdeltaT, we can calculate the quantity of heat that is transferred from the iron to the ice. The change in temperature (deltaT) for the iron is 80 degrees C (since it goes from 80 degrees C to 0 degrees C), and for the ice, it is 0 degrees C (since it goes from 0 degrees C to 0 degrees C).

Thus, we have Q = (0.11 cal/g*C)(50 g)(80 C) = 440 cal.

We also know that it takes 80 calories to melt 1 gram of ice. Therefore, the mass of ice that will melt can be calculated as:

440 cal = (80 cal/g)(x g)

Solving for x, we get x = 5.5 g. Therefore, 5.5 grams of ice will melt when the 50 g chunk of iron at 80 degrees C is dropped into the cavity in the ice at 0 degrees C.

I hope this explanation helps. Please let me know if you have any further questions.

## 1. What is "Thermo Energy"?

Thermo Energy is a term used to describe the energy that is present in a system due to its temperature.

## 2. How does iron and ice interact to produce thermo energy?

Iron and ice interact to produce thermo energy through the process of thermal expansion. When heated, the iron expands and exerts pressure on the surrounding ice, causing it to melt and release thermal energy.

## 3. Can thermo energy be harnessed and used for practical purposes?

Yes, thermo energy can be harnessed and used for practical purposes, such as in power plants where heat is used to turn water into steam to generate electricity.

## 4. Is thermo energy a renewable or non-renewable resource?

Thermo energy is a non-renewable resource, as it is derived from the finite amount of heat stored in a system.

## 5. What are some common applications of thermo energy?

Some common applications of thermo energy include heating and cooling systems, power generation, and industrial processes such as metal smelting and food processing.

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