# Measuring specific heat capacity

• Janinever
In summary, the textbook says that a 0.040 kg mass of unknown material has a heat capacity of 1300 J/(kg.K), but when I try to calculate it using the equation provided in the book, I get 1656.266667 J/(kg.K). I'm not sure how to fix this mistake, or why the textbook would give a different answer than the calculator.
Janinever
I completed high school 9 years ago... please bare with me :)

My problem is with how they calculate the actual answer - this is from an example problem in my textbook.

## Homework Statement

A calorimeter cup is made from 0.15kg of Alu and contains 0.20 kg of water. Initially the water and cup have a common temp of 18C. A 0.040 kg mass of unknown material is heated to a temp of 97C and then added to the water. The temp of the water, the cup and unknown material is 22C after thermal equilibriam is reestablished. Ignoring small amount of heat gained by thermometer, find the heat capacity of the unknown material.

## Homework Equations

Using the equation that they've explained in the book and taking all the data and putting it in the equation (exactly as in my book) see below :

## The Attempt at a Solution

[9.00 x 10^2 J/(kg.C](0.15kg)(4.0C) + [4186 J/(kg.C)](0.20kg)(4.0C)
--------------------------------------------------------------------
(0.040kg)(75.0C)

The answer in the book is 1300 J/(kg.C)

With the calculator I get 1656.266667

So I have no clue how they got to 1300?

Please could someone help me understand how they get to the answer? I understand the rest.

Thanks so much!

hi Janinever, your mistkae is here. you had used specific heat with kelvin not in degree celcuis in this cade you should use the temperture with kelvin also or change the specifi heat constant with degre 9.102 J.Kg-1.°K-1 it's in kelvin k=273+T in(degree celcuis)

Thanks for the reply :) don't understand 100%? :)

i mean pay attention in units, look in this table:

http://data.imagup.com/10/1146563447.jpg

you should change the tempertaure to kelvin or recalculate the specific heat from J/(kg.K to J/(kg.C) you know the temperature in kelvin=273+the temperature in degree celcuis.
.

The entire equation from start to finish is in Celsius - so I'm not sure how converting to K is going to help when the books final answer is in C? Theres no mention or indication of k anywhere in this equation? Could you explain in detail what you mean maybe? Or re read my question just incase we are misunderstanding each other somewhere :)

Janinever said:
I completed high school 9 years ago... please bare with me :)

My problem is with how they calculate the actual answer - this is from an example problem in my textbook.

## Homework Statement

A calorimeter cup is made from 0.15kg of Alu and contains 0.20 kg of water. Initially the water and cup have a common temp of 18C. A 0.040 kg mass of unknown material is heated to a temp of 97C and then added to the water. The temp of the water, the cup and unknown material is 22C after thermal equilibriam is reestablished. Ignoring small amount of heat gained by thermometer, find the heat capacity of the unknown material.

## Homework Equations

Using the equation that they've explained in the book and taking all the data and putting it in the equation (exactly as in my book) see below :

## The Attempt at a Solution

[9.00 x 10^2 J/(kg.C](0.15kg)(4.0C) + [4186 J/(kg.C)](0.20kg)(4.0C)
--------------------------------------------------------------------
(0.040kg)(75.0C)

The answer in the book is 1300 J/(kg.C)

With the calculator I get 1656.266667

So I have no clue how they got to 1300?

Please could someone help me understand how they get to the answer? I understand the rest.

Thanks so much!

Could be you're having finger problems with the calculator If I run the same calculation as stated above I obtain 1296 J*kg-1*K-1, which rounds nicely to 1300 J*kg-1*K-1.

Then I'm definitely doing something wrong because I get the same answer over and over :( Will sit with it a while longer and see if I manage to find the problem. Thank you

Got it! :) Thanks Gneil! I haven't used a scientific calculator in years lol so that explains my horrible mistake with the exponent! Thank you SO MUCH!

## What is specific heat capacity?

Specific heat capacity is the amount of heat required to raise the temperature of one unit of mass of a substance by one degree Celsius.

## Why is measuring specific heat capacity important?

Measuring specific heat capacity is important because it helps us understand how different materials respond to heat and how much energy is needed to change their temperature. This information is crucial for various scientific and engineering applications, such as designing heating and cooling systems.

## How is specific heat capacity measured?

Specific heat capacity is measured by conducting an experiment where a known amount of heat is added to a substance and its temperature change is recorded. The equation q = mcΔT is used to calculate specific heat capacity, where q is the heat added, m is the mass of the substance, and ΔT is the change in temperature.

## What factors can affect the measurement of specific heat capacity?

The accuracy of the measurement of specific heat capacity can be affected by several factors, such as the purity and density of the substance, the method of heating, and the accuracy of the temperature measurement.

## What are some common units for specific heat capacity?

The most commonly used units for specific heat capacity are joules per gram per degree Celsius (J/g°C) and calories per gram per degree Celsius (cal/g°C). However, other units such as joules per kilogram per Kelvin (J/kgK) and British thermal units per pound per degree Fahrenheit (BTU/lb°F) may also be used.

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