Q Lost Q Gained
Q=mc∆T q gained = q lost q = -q mc∆T = - mc∆T 1. The specific heat of ethanol is 2.46 J/goC. Find the heat energy required to raise the temperature of 193 g of ethanol from 19 oC to 35 C. When a 120 g sample of aluminum absorbs 9612 J of heat energy, its temperature increases from.
I've been playing this game for a long time now, and from experience, try to be AT LEAST lv. 70 when going to the Temple of Enlightenment. For the final boss, you can fight him around lv. 50-55 but I recommend that you beat side-quests and get some items like Quad-Element Ameulet, that way you'll be around lv. 53 and have some good items to crush him with. The q lost by one object will be the q gained by the other object. Lost = q gained. Continuing with the H. 2 from the earlier problemThe hydrogen was put in 9.179 moles of a substance at 274.4 K and it heated to 280.4 K, what was the substance? Lost = 4147.2 J4147.2 = 9.179 mol C (280-274K).
When an object is heated, its temperature increases and when the object cools, its temperature decreases.
Heat capacity (C) is the amount of heat (q) required to raise the temperature of an object one degree Celsius.
The units for heat capacity are J/oC.
(The unit is read as Joules per degree Celsius).
The equation which describes this relationship is:
C = q/DT
Q Lost Q Gained
C = is the heat capacity of the object
q = is the amount of heat entering or leaving the object
The change in temperature (DT) of the object is defined as: DT = Tfinal - Tinitial
A positive value for heat (+q) means that heat is entering the object. In this case, the final temperature of the object
will be higher than the initial temperature.
lower than the initial temperature.
Measurements Of Heat
| EXAMPLE 1 How much heat is required to raise the temperature of an object from 25.00oC to 60.00oC, if the heat capacity of the object is 0.755 J/oC? C = q/DT 0.755 J/oC = q / (60.00oC - 25.00oC) q = +26.426 J |
| EXAMPLE 2 An object with a heat capacity of 38.25 J/oC lost 4.593 kJ of heat. If the initial temperature was 235.19 oC, calculate the final temperature after the cooling process. C = q/DT 38.25 J/oC = (-4.593 kJ)(1000 J/kJ) / ( Tfinal – 235.19oC) Tfinal = 115.11 oC |
Heat Q
Specific heat capacity (or simply specific heat) is the amount of heat required to raise the temperature of one gram of a
substance one degree Celsius.
The equation which describes this relationship is:
q = (mass of the substance)(specific heat of the substance)(DT)| EXAMPLE 3 What amount of heat is released when a 100.0 gram sample of copper cools from 95.00°C to 65.00°C? The specific heat capacity of copper is 0.382 J/g×°C. q = (mass) (specific heat) (DT) = (100.0 g) (0.382 J/g×°C) (65.00°C - 95.00°C) = -1146 J |
Calorimeter
A calorimeter is a device used to measure heat changes that accompany physical or chemical processes.
Exothermic processes carried out inside a calorimeter result with an increase of the calorimeter temperature.
Endothermic processes carried out inside a calorimeter result with a decrease of the calorimeter temperature.
The heat given off in an exothermic process will be the same amount of heat gained by the calorimeter.
The heat absorbed in an endothermic process is the same amount of heat lost by the calorimeter.
The heat change of the process is therefore equal to but opposite in sign to the heat change of the calorimeter.
- qprocess = qcal| EXAMPLE 4 A 123.57 g sample of iron was heated to 100.00oC and then placed in a calorimeter with a heat capacity of 209.2 J/oC. The temperature of the calorimeter rose from 20.00oC to 36.80oC. Calculate the specific heat capacity of iron. heat gained by calorimeter = C(DT) = (209.2 J/oC)(36.80oC – 20.00oC) = +3514.56 J -(heat lost by iron) = heat gained by calorimeter heat lost by iron = (mass)(specific heat capacity)(DT) -3514.56 J = (123.57 g)(specific heat capacity)(36.80oC – 100.00oC) specific heat capacity = 0.450 J/g×°C |
When a reaction is carried out inside a calorimeter, the heat lost or absorbed by the reaction (qrxn) is equal to but
opposite in sign to the heat gained or lost by the calorimeter (qcal).
| EXAMPLE 5 A certain calorimeter has a heat capacity of 8.952 kJ/oC. When a 3.500 g sample of carbon was burned to CO2 in the calorimeter, the temperature of the calorimeter rose from 25.00 oC to 38.18 oC. What is the heat of combustion of carbon in joules per gram? Heat gained by calorimeter (qcal) = C(DT) qcal = (8.952 kJ/oC)(38.18 oC - 25.00 oC) = +117.99 kJ -(heat produced by the rxn) = heat gained by calorimeter -qrxn = qcal qrxn = -117.99 kJ = -117990. J heat per gram = J/g = -117990. J/3.500 g = -3.371 x 104 J/g |
THERMOCHEMICAL EQUATIONS
The heat flow for a reaction at constant pressure, qp, is called enthalpy, DH.
Exothermic reactions have negative enthalpy values (-DH).
Endothermic reactions have positive enthalpy values (+DH).
An equation which shows both mass and heat relationships between products and reactants is called a thermochemical equation.
2 H2(g) + O2(g) ----> 2 H2O(l)DH = -571.6 kJ
The magnitude of DH is directly proportional to the amount of reactants or products.
A + 2 B ----> C DH = -100 kJ
1/2 A + B ----> 1/2 C DH = -50 kJ
DH for a reaction is equal in magnitude but opposite in sign for the reverse reaction.
A + 2 B ----> C DH = -100 kJ
Measuring Specific Heat
C ----> A + 2 B DH = +100 kJQ Gained Equals Q Lost
| EXAMPLE 6 Based upon the thermochemical equation given, calculate the heat associated with the decomposition of 1.15 g of NO2(g). N2(g) + 2 O2(g) ----> 2 NO2(g)DH = +67.6 kJ A. Calculate moles of NO2(g). B. Reverse the thermochemical equation and change sign of enthalpy. C. Calculate heat. |