physics
posted by tasha .
A 1.3 Kg block of ice is initially at a temp of 5 degrees celsius. If 7.5 *10^5 J of heat are added to the ice, what is the final temperature of the system?
I multiplied 1.3 Kg of ice by the specific heat of ice (2090), then I divided by the amount of heat added. I came up with 0.0036 1/K, so I took the inverse and found it to be 276 K, which is equivalent to 4.78 degrees Celsius. This is not right. Can you help me?
Thank you.
Well, you did just about everything wrong.
First, see if the ice warms up to the melting point.
Heattomelt=specificheat*deltaTem[
and the deltatemp to zero C is 5
Heattomelt=2090J/kg*5=1.046E4
So there is left over heat when it gets to OC.
Leftover heat= 75E4 j1.046E5 J=74E4
Heat required to melt ice=massice*latent heat fusion
= 1.3kg*3.34E5 J=43.4E4Joules
So, all the ice melts, and there is still leftover heat.
Heat left over= 74E443.4E4 Joules.
Now find out the final temp...
74E443.4E4=1.3kg*4.18kJ/kg*deltaTemp
or solve for deltaTemp. I get a finalt emp of nearly sixty degrees C.
I will be happy to critique your thinking.
Well, you did just about everything wrong.
First, see if the ice warms up to the melting point.
Heattomelt=specificheat*deltaTem[
and the deltatemp to zero C is 5
Heattomelt=2090J/kg*5=1.046E4
So there is left over heat when it gets to OC.
Leftover heat= 75E4 j1.046E5 J=74E4
Heat required to melt ice=massice*latent heat fusion
= 1.3kg*3.34E5 J=43.4E4Joules
So, all the ice melts, and there is still leftover heat.
Heat left over= 74E443.4E4 Joules.
Now find out the final temp...
74E443.4E4=1.3kg*4.18kJ/kg*deltaTemp
or solve for deltaTemp. I get a finalt emp of nearly sixty degrees C.
I will be happy to critique your thinking.
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