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When 418 joules of heat energy are added to 10 grams of water at 20C the final temperature of the water will be?
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What metal is the highest final temperature after the heat energy has been added?
The metal with the lowest thermal capacity.
If 980kJ of energy are added to 6.2L of water at 291K what will the final temperature of the water be?
980000/ (6200*4.180)+18`c or 291K final answer should be 56`c or 329K.
What effect does the initial temperature of the water have on change in temperature of the water after the hot metal is added?
When hot metal is added into the water then the metal looses its energy into the water and this heat is gained by the water, so the temperature gets increases when hot metal added into it i.e final temperature is greater than initial temperature of water.
How 750 calories of heat are added to 250 grams of Lead c equals 0.031 cal g C initially at a temperature of 28.0 C What will be the final temperature of this piece of Lead?
I need to make some conversions for my own convenience. 750 calories (4.184 Joules/1 calorie) = 3138 Joules ----------------------- Lead specific heat (c) = 0.160 J/gC Now, q = mass * specific heat * change in temperature 3138 Joules = (250 g Pb)(0.160 J/gC)(Tf - 28.0 C) 3138 J = 40Tf- 1120 4258 = 40Tf 106 Celsius = Final temperature of lead ----------------------------------------------------
What quantity is measured by calorimeter?
Final Temperature Initial Temperature Specific Heat Capacity of Calorimeter Plug the values into the equation: q = C( Tf - Ti ) , where C = specific heat, Tf = Final Temperature, and Ti = Initial Temperature.
How do you calculate the required energy for rising the tempareture from x to y of m mass of steel?
q( enthalpy of heat in Joules-energy ) = Mass of steel * heat capacity of steel * ( Temperature final - temperature initial )
What are the energy formulas involving specific heat mass and temperature change?
E = mass x sp ht x Δ°t (Finding Energy)where E (Energy) or Q (Quantity of Heat), mass (g), sp ht (aka specific heat, J/g°C*[typical] or cal/g°C or kcal/g°C), and Δ°t (temperature change). Finding Temperature ChangeDivide energy by mass multiplied by specific heat. Δ°t = Energy-- Mass * sp htIn order to find the final temperature (if problem is asking for this), add or subtract the original temperature and the new temperature together.Tf = original temperature +/- new temperatureIf energy is added, the temperatures will be added together; if energy is removed, the temperatures will be subtracted.Finding MassDivide energy by specific heat multiplied by temperature change. Mass = Energy------- sp ht * Δ°tFinding Specific HeatDivide energy by mass multiplied by temperature change. Sp ht = Energy------- Mass * Δ°tConverting Form of Energy (joules, kcal, and cal)Sometimes a problem will have E be shown in cal/g°C or kcal/g°C but will be asking for Joules or even vice versa. This means a conversion has to take place. Cal --> Joules and Joules --> cal-Calories (Cal) --> Joules (J)Multiply # cal by 4.184 Joules (J).Conversion Factor# cal x 4.184 J = Joules---------- 1 cal1 cal = 4.184 Joules-Joules (J) --> Calories (Cal)Divide # Joules (J) by 4.184Conversion Factor# J x 1 cal = cal--- 4.184 J1 Joule = 0.239005736 calKcal --> Joules (J) and Joules (J) --> Kcal-Joules --> kcal (Joules --> cal --> kcal)Divide # J by # kcal multiplied by 103Conversion Factor# Joules x 1 cal - * - 1kcal = kcal---------- 4.184J -- 103 cal1 kcal = 4,184 Joules 1 Joule = 0.000239005736 kcal-Kcal --> Joules (J) (Kcal --> cal --> Joules)Multiply # kcal by 103 cal by 4.184JConversion Factor# kcal * 103 cal * 4.184J = Joules (J)----------- 1 kcal --- 1 calAnother relationship that is good to understand: 1 kcal = 1000 (103) cal 1 cal = 0.001 (10-3) kcal
What metal is the highest final temperature after the heat energy has been added?
The metal with the lowest thermal capacity.
What is the final temperature if 4.00 x 104 joules of energy are removed from 500.0g H2O at 10.0 degrees Celsius?
E = mass x sp ht x Δ°tIn order to calculate the final temperature change, we must also find the temperature change.1. Find the temperature change.Divide joules by (mass x specific heat).Δ°t = 40,000J/500.0g x 4.184J/g°CΔ°t = 19.1°C2. Calculate the final temperature.Tf = 10.0 - 19.1Tf = -9.1°CThere was a decrease in temperature, indicating that it was an endothermic reaction (as energy was removed as stated in the question).
What substance is used as the standard against which you measure specific heats?
If you can measure 3 of these 4 things then you can use this formula q( energy in Joules ) = Mass * specific heat * temperature final - temperature initial
If 980kJ of energy are added to 6.2L of water at 291K what will the final temperature of the water be?
980000/ (6200*4.180)+18`c or 291K final answer should be 56`c or 329K.
What will the final temperature be if 688 calories of heat is released by 25 mL of water with an initial temperature of 80 degrees celsius?
Do some converting first. 688 calories (4.184 Joules/1 calorie) = 2878.592 Joules 25 ml of water = 25 grams q(Jolules) = mass * specific heat * (Temp. final - Temp. initial) 2878.592 Joules = 25 grams Water * 4.180 J/gC * (Temp Final - 80C ) 2878.592 Joiles = 104.5( Temp. Final) - 8360 11238.592 =104.5(Temp. Final) 107.55 Celsius Final Temperature ( call it 108 C )
What effect does the initial temperature of the water have on change in temperature of the water after the hot metal is added?
When hot metal is added into the water then the metal looses its energy into the water and this heat is gained by the water, so the temperature gets increases when hot metal added into it i.e final temperature is greater than initial temperature of water.
How 750 calories of heat are added to 250 grams of Lead c equals 0.031 cal g C initially at a temperature of 28.0 C What will be the final temperature of this piece of Lead?
I need to make some conversions for my own convenience. 750 calories (4.184 Joules/1 calorie) = 3138 Joules ----------------------- Lead specific heat (c) = 0.160 J/gC Now, q = mass * specific heat * change in temperature 3138 Joules = (250 g Pb)(0.160 J/gC)(Tf - 28.0 C) 3138 J = 40Tf- 1120 4258 = 40Tf 106 Celsius = Final temperature of lead ----------------------------------------------------
How many joules of energy are required to raise the temperature of 150.0g of water from 15.0 degrees C to 45.0 degrees C?
The specific heat capacity of water is 4186 joules per kilogram. That is to raise 1kg or 1 litre of water by 1 degree you will need to add 4186 joules of energy. So for 15grams over 25 degrees you will need 4186/1000*15*15 joules.
What happens to the energy when the current goes through the wires?
The current does work (W) on the wire, causing its internal energy (U) to increase. This, in turn, raises the temperature of the wire above its surroundings, so heat transfer (Q) takes place away from the wire. This is summarised as follows:(W - Q) = U = m c (Tf - Ti)where:W = work (joules)Q = heat (joules)m = mass of wire (kilograms)c = specific heat capacity of wire (joules per kilogram kelvin)Tf = final temperature (kelvin)Ti = initial temperature (kelvin)
What quantity is measured by calorimeter?
Final Temperature Initial Temperature Specific Heat Capacity of Calorimeter Plug the values into the equation: q = C( Tf - Ti ) , where C = specific heat, Tf = Final Temperature, and Ti = Initial Temperature.
