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Energy transferred (W) is given by
where
So in this case,
Therefore the final temperature of the water will be
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How do you calculate the energy required to produce 7.00 mol $Cl_2O_7$ on the basis of the following balanced equation: $2Cl_2(g) + 7O_2(g) + 130 kcal -> 2Cl_2O_7(g)$?
You gots... $2Cl_2(g) + 7O_2(g) + 130*kcal rarr 2Cl_2O_7(g)$ Another way of writing this is as... $2Cl_2(g) + 7O_2(g) rarr 2Cl_2O_7(g)$ $DeltaH_"rxn"^@=+130*kcal*mol^-1$ And when we write $mol^-1$ we mean per...
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What is the initial pressure of a gas having an initial temperature of 90.5 K, an initial volume of 40.3 L, a final pressure of 0.83 atm, a final temperature of 0.54 K and a final volume of 2.7 L?
$PV = nRT$ $(PV)/T = nR$ $therefore (P_1V_1)/T_1 = (P_2V_2)/T_2$, the . $(P_1*"40.3 L")/("90.5 K")=("0.83 atm"*"2.7 L")/("0.54 K")$ $P_1 approx 9.3atm$
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What is the initial pressure of a gas having an initial temperature of 90.5K, an initial volume of 40.3 L, final pressure of .83 atm, a final temperature of .54 K and a final volume of 2.7L?
Use the . $(P_1V_1)/T_1=(P_2V_2)/T_2$ Given $V_1="40.3 L"$ $T_1="90.5 K"$ $P_2="0.83 atm"$ $V_2="2.7 L"$ $T_2="0.54 K"$ Unknown $P_1$ Equation $(P_1V_1)/T_1=(P_2V_2)/T_2$ Solution Rearrange the equation to isolate $P_1$ and solve. $P_1=(P_2V_2T_1)/(T_2V_1)$ $P_1=((0.83"atm"...
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How much energy must be transferred to raise the temperature of a cup of coffee (250 mL) from 20.5 C (293.7 K) to 95.6 C (368.8 K)? Assume that water and coffee have the same density (1.00 g/ mL), and specific heat capacity 4.184 J/ gK)?
The energy required to heat an object is given by the formula $color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "$ where $q$ is the energy required $m$ is the mass...
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Two 20.0-g ice cubes at -11.0 °C are placed into 265 g of water at 25.0 C. Assuming no energy is transferred to or from the surroundings, how do you calculate the final temperature of the water after all the ice melts?
Let $T_f$ be the final temperature of mixture. The ice absorbs heat during two process $\text{Ice}\to\text{cooling from}-11^\circ C \ \text{to }0^\circ C \to \text{melting at }0^\circ C\to \text{cooling from }0^\circ...
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A 60 g piece of metal at an initial temperature of 92°C is transferred into 70 g of water initially at 24.5°C. The final temperature of the water and the metal is 40°C. How do you determine the specific heat of the metal?
$q_"gain" = q_"lost"$ $(60g)(92^o - 40^oC)(x) = (70g)(24.5^oC - 40^oC)(4.18)"$ 4.18 is specific of water. Solve for x $3120x=-4535.3$ $x = 4535.3/3120 or 45353/31200$ $x = -1.45362$ This is the...
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How much boiling water would you need to raise the bath to body temperature (about 37 ∘C)? Assume that no heat is transferred to the surrounding environment.
Express your answer to two significant figures and include the appropriate units.
The idea here is that the heat given off by the boiling water will be equal to the heat absorbed by the room-temperature sample. $color(blue)(ul(color(black)(q_"absorbed" = -q_"given off")))" "...
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There are two cups filled with equal amount of tea and coffee. A spoonful of
coffee is first transferred from coffee-cup to the tea-cup and then a spoonful
from the tea-cup is transferred to the coffee-cup, then?
The assumptions I will make are: The spoonfuls transferred are of the same size. The tea and coffee in the cups are incompressible fluids which do...
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When a body is projected vertically up from the ground, its potential energy and kinetic energy at a point P are in the ratio 1:5. if the same body is projected up with half the previous velocity then at the same point P the ratio will be what?
$KE=1/2mv^2$ $PE=mgh$ @$P$: $(1/2mv^2)/(mgh)=5$ $:.(v^2)/(2gh)=5 " "color(red)((1))$ The velocity is now halved but $h$ remains the same so we can write: $((v/2)^2)/(2gh)=R$ Where $R$ is the ratio of $(KE)/(PE)$ $:.((v^2)/(4))/(2gh)=R$...
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Describe the energy conversion used in the following scenarios:
A. Natural gas is used to heat water.
B. rubbing your hand together on a cold day
C. A car (heat) engine
D. Your body temperature and the ability to do work
E.water runs over a water fall?
A. This looks like chemical potential becoming kinetic energy in the molecules of water (or as a simpler answer you might just say heat in the water) B. This is...
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