What amount of energy is required to melt a 10.7 g piece of ice at 0 C? The heat of fusion of ice = 333 Jg-1

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

We perform the physical reaction........

$H_2O(s) +Deltararr H_2O(l)$; both reactants and products are at $0$ $""^@C$, and thus we need only assess the energy transfer of the PHASE change........

We are given the $"heat of fusion"$ of ice in $J*g^-1$....

We gots a $10.7*g$ mass of ice, and so we take the product.........

$10.7*cancelgxx333*J*cancel(g^-1)xx10^-3*J*kJ^-1=??J$....

And this represents HEAT, $Delta$, put into the system.........is the sign correct?

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
How much heat is required to convert 20.0 g of ice at -50.0°C to liquid water at 0.0°C? The specific heat of ice is 2.06 J/(g *$"^o$C) and the heat of fusion of water is 334 J/g?

We interrogate the process..... $H_2O(s)stackrel(Delta_1, -20^@C...0^@C)rarrH_2O(s)stackrel(Delta_2=Delta_"fusion")rarrH_2O(l)$ Now $Delta_1=20.0*gxx2.06*J*g^-1*""^@C^-1xxDeltaT=$ $20.0*gxx2.06*J*g^-1*""^@C^-1xx50$ $""^@C=2060*J$ And $Delta_2=20.0*gxx334*J*g^-1=$ $20.0*gxx334*J*g^-1=6680*J$ $"Total heat"=Delta_1+Delta_2=2060*J+6680*J=??*J$

1 Answers 1 views
Given the Following: heat of fusion of ice = 80.0 cal/g; specific heat of ice = 0.480 cal/g; and specific heat of water = 1.00 cal/g. Calculate the amount of heat required to melt 30.0 g of ice at 0 degrees celsius?

$H_2O(s) +DeltararrH_2O(l)$....... Note that BOTH the ice and water have a specified temperature of $0$ $""^@C$, and thus we assess the heat involved in the phase change, and thus we...

1 Answers 1 views
The heat of fusion for water is 80. cal/g. How many calories of heat are needed to melt a 35 g ice cube that has a temperature of 0 C?

The formula for the heat required to melt a solid is $color(blue)(|bar(ul(color(white)(a/a) q= mΔ_"fus"H^°color(white)(a/a)|)))" "$ where $q = "the amount of heat"$ $m = "the mass...

1 Answers 1 views
What is the minimum amount of heat required to completely melt 20.0 grams of ice at its melting point? A. 20.0J B. 83.6J C. 6680J D. 45,200J How do you solve

In order to be able to answer this question, you need to know the value of water's of fusion, $DeltaH_f$. As you know, a substance's enthalpy of fusion tells...

1 Answers 1 views
What amount of heat is required to melt $"300 g"$ of ice at its freezing point if the enthalpy of fusion is $"6.02 kJ/mol"$?

For a phase transition, we assume a constant temperature (and also, pressure). So, the heat flow $q$ into a substance is equal to the change $DeltaH$. $bb(q_P)$ $= bb(mDeltaH)$,...

1 Answers 1 views
A35.8 g sample of cadmium metal was melted by an electric heater providing 4.66 J/s of heat. If it took 6.92 min from the time the metal began to melt until it was completely melted, what is the heat of fusion per mole of cadmium?

This is a pretty creative problem. $35.8g * (mol)/(112.4g) = 0.319mol$ of cadmium are melted. $6.92min * (60s)/(min) * (4.66J)/s * (kJ)/(10^3J) approx 1.93kJ$ of heat were required, hence,...

1 Answers 1 views
What is the total amount of heat required to completely melt 347 grams of ice at its melting point?

We interrogate the physical change...... $H_2O(s) +Deltararr H_2O(l)$ Clearly this process is ENDOTHERMIC, and that the latent heat of fusion for ice is $334*kJ*kg^-1$. We have a mass of $347*g$...

1 Answers 1 views
The latent heat of fusion for ice is 6.0 kJ/mole. In order to melt 36 g of ice (solid $H_2O$) at 0°C, how much energy is required?

The molar latent heat of fusion, which is an alternative name given to the of fusion, tells you how much heat is required in order to convert a specific...

1 Answers 1 views
As a 6 kg of a liquid substance at its freezing point completely freezes , it gives enough heat to melt 3 kg of ice at $0^o$ C the heat of fusion of the substance is ?

The of fusion, $DeltaH_"fus"$, tells you the energy needed to convert $"1 g"$ of a given substance from solid to liquid at its melting point. Similarly, you can...

1 Answers 1 views
The heat of fusion of water is 336 J/g. What is the molar heat of fusion (j/mol) of water?

Use the fact that one mole of water is equal to $18.02$ grams of water: $((336J)/(1cancel(gH_2O)))((18.02cancel(gH_2O))/(1molH_2O)) = 6050J/(mol)$ A more conventional unit for of fusion, enthalpy of vaporization, etc. is...

1 Answers 1 views
X