Measurement of Heat, Heat capacity and Specific Heat capacity

Measurement of Heat Energy

Heat is a form of energy called thermal energy that flows due to temperature difference. It is measured in joules.

The quantity of heat Q received by a body is proportional to its mass (m), and temperature change (0₂ –  0₁) and on the nature of the material the body is made of.

Thus Q∝(θ2−θ1)Q=MC(θ2−θ1)

C is a constant of proportionality called the specific heat capacity of the body, which depends on the nature of the body.

The quantity of heat energy possessed by a body depends on these three quantities:

(i) the change in temperature (θ₂ – θ₁)

(ii) the specific heat capacity of the body (C)

(iii) mass of the body (m)

Heat Capacity

This is the quantity of heat required to raise the temperature of the entire body by one degree rise in temperature(1k).  It is measured in Joules/K.

H = MC

Specific Heat Capacity

Click on the slideshow below to learn more

https://www.slideshare.net/slideshow/embed_code/key/M0vsXIoDxOOIoT4.2 Specific Heat Capacity from Nur Farizan

Specific heat capacity of a substance is the heat required to raise the temperature of a unit mass(1kg)  of the substance through a degree rise in temperature(1⁰C or 1K)

Q=MC(θ2−θ1)

Q is quantity of heat,  C is specific heat capacity,  (θ2−θ1) change in temperature and m is the mass of the substance.

C=QM(θ2−θ1)

The unit is JKg^-1K^-1

Example 1:

What is meant by the statement: The specific heat capacity of copper is 400Jkg^-1k^-1?

Solution:

It means the quantity of heat required to raise the temperature of a unit mass of copper through one degree rise in temperature is 400J

Example 2:

How much heat is given out when a piece  of iron of mass 50g and specific heat capacity 460Jkg^-1K^-1 cools from to 25⁰C.

Solution:

M=50g=0.05kg,=460Jkg−1k−1,θ2−θ1=85−25=60oCQ=MC(θ2−θ1)Q=0.05×460×60=1380J

Determination of the Specific Heat Capacity of a Solid by Method of Mixtures

The solid lead block is weighed on a balance to be M₁.  A lagged calorimeter is dried and weighed to be M₂.  It is then reweighed to be M₃ when half filled with water.  The initial temperature of the water is taken to be 0₁.

The lead block is suspended in boiling water with a temperature 0₂ after which it is transferred to the calorimeter and the mixture stirred to maintain a uniform temperature 0₃.

The specific heat capacity of the lead can be calculated using the fact that heat loss by the lead = heat gained by calorimeter and water.  Given the specific heat capacity of calorimeter and water to be C_c and C_w respectively.

M1Cc(θ2−θ3)=M2Cc(θ3−θ1)+(M3−M2)Cw((θ3−θ1))C1=(M2Cc)(θ3−θ1)+(M3−M2)Cw(θ3−θ1)M1(θ2−θ3)

Precautions:

1. The calorimeter should be well lagged.
2. The mixture should be well stirred to ensure even distribution of heat.
3. The hot solid should be quickly transferred to prevent loss of heat.

Calculations Using Method of Mixtures

Example 3:

An iron rod of mass 2kg and at a temperature of 280⁰C is dropped into some quantity of water initially at a temperature of 30⁰C . If the temperature of the mixture is 70⁰C ,  calculate the mass of the water.   [Neglect heat losses to the surroundings.] [Specific heat capacity of iron = 460Jkg^-1K^-1_, Specific heat capacity of water = 4200Jkg^-1K^–]

Solution:

Mass of iron rod = M₁ = 2kg

Temperature of hot iron rod = θ₂ = 280⁰C

Initial temperature of water  = θ₁ = 30^oC

Final temperature of mixture = θ₃ = 70⁰

Specific heat capacity of iron = C_{I  =} 460Jkg^-1K^-1

Mass of water = M₂ = ?

Heat lost by hot iron = heat gained by water

M1CI(θ2−θ3)=M2CW(θ3−θ1)M2=M1CI(θ2−θ3)CW(θ3−θ1)M2=2×460×(280−70)4200(70−30)M2=2×460×2104200×40M2=193200168000=1.15kg

Example 4:

A piece of copper of mass 120g is heated in an enclosure to a temperature of 125⁰C. it is then taken out of the enclosure and held in air for half a minute and dropped carefully into a copper calorimeter of mass 105g containing 200g of water at 20⁰C. The temperature of the water rises to 25⁰C. Calculate the rate at which heat is being lost from the piece of copper when it is held in air. (specific heat capacity of water is 4200Jkg ^{– 1 0}C ^{– 1} ,  specific heat capacity of copper is 400J kg ^{– 10}C ^{– 1}       WAEC)

Solution:

θ₁ = 125⁰C,  θ₂ = 25⁰C,  mass of copper (M_c) = 120g = (120/1000)kg = 0.12kg

Heat lost by copper =McCc(θ2−θ1)=0.12×400×(125−25)=480J

Mass of calorimeter (m_c) =105g=(1051000)kg=0.105kg

Specific heat capacity of calorimeter (c_c) =400Jkg−10C−1

Mass of water =(2001000)kg=0.2kg

Change in temperature Δθ=(25−20)oC

Heat gained by calorimeter and water

=McCcΔθ+MwCwΔθ=0.105×400×(25−20)+0.2×4200×(25−20)=4410J

Heat lost to air =4800−4410=390J

Therefore, rate of lost of heat to air

=39030=13Js−1

Determination of Specific Heat Capacity of a Solid by Electrical Method

To calculate the specific heat capacity C_b of a solid brass block, we make two holes in a weighed brass block into which a thermometer and a heating element connected to a source of power supply are inserted.  Oil is poured in the holes to ensure thermal conductivity.  Assuming no heat is lost to the surrounding, the total amount of electrical heat energy supplied by the coil, Ivt = heat gained by the brass, MC_b0

Ivt=MCbθ ——(1)

From v = IR (Ohms law)

I2Rt=MCbθ ——(2)

v2tR=MCbθ ——(3)

Example 5:

A liquid of specific heat capacity 3Jg ^{– 1} k ^{– 1} rises from 15⁰C to 65⁰C in one minute when an electric heater is used. If the heater generates 63KJ per minute, calculate the mass of the liquid.

Solution:

Specific heat capacity of liquid

C1=3Jg−1k−1=3000Jkg−1k−1Δθ=65−15=50oC

Heat supplied by heater = heat gained by water

Ivt = M_l x C_l Δθ where M_l = mass of liquid

63000=M1×3000×50M1=630003000×50=0.42kg

Example 6: A certain metal of mass 1.5kg at initial temperature of 27⁰C, absorb heat from electric heater of 75W rating for 4 minutes. If the final temperature was 47⁰C, calculate the specific heat capacity of the metal and its heat capacity.

Solution:

Time ‘t’ = 4 minutes = 4 × 60 = 240s. power IV = 75W,  mass of metal ‘m’ = 1.5kg.

Heat supplied by electric heater = heat gained by the metal

IVt=mcΔθ75×240=1.5×c×(47–27)75×240=1.5×c×20c=75×2401.5×20=600Jkg−1K−1

Heat capacity =mc=1.5×600=900JK−1

GENERAL EVALUATION ( POST YOUR ANSWER IN THE COMMENT BOX AT THE BOTTOM  FOR RATING AND DISCUSSION)

1. 250g of lead at 170⁰C is dropped into 100g of water at 0⁰ If the final steady temperature is 12⁰C, calculate the specific heat capacity of lead.  (C_w= 4.2 x 10³ J/kgk)
2. A 2000W electric heater is used to heat a metal object of mass 5kg initially at . If a temperature rise  of  is obtained after 10min, calculate the heat capacity of the material.

Determination of Specific Heat Capacity of Liquid by Electrical Method

Apparatus:

Thermometer, liquid, calorimeter, heater, stop clock, chemical/beam balance

Method:

A calorimeter of known heat capacity (M_cC_c) is used and a known mass of liquid( M₁)  is placed in the calorimeter, the temperature of the liquid is recorded (θ₁₎). The known quantity of heat (VIt) is  recorded by taking readings from the voltmeter, ammeter and stop watch. The final temperature is recorded (θ₂).

Calculations:

Electrical energy supplied by the heater = Heat energy absorbed by the calorimeter and water.

VIt=M1CL(θ2−θ1)+McCc(θ2−θ1)CL=VIt−McCc(θ2−θ1)C1(θ2−θ1)