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 j-1.046E5 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= 74E4-43.4E4 Joules.
Now find out the final temp...
74E4-43.4E4=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 j-1.046E5 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= 74E4-43.4E4 Joules.
Now find out the final temp...
74E4-43.4E4=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.
To solve this problem correctly, you need to consider the different stages of energy transfer that occur when adding heat to the ice. Here's a step-by-step explanation of how to solve it:
1. Determine the heat required to warm up the ice from -5 degrees Celsius to its melting point at 0 degrees Celsius. The formula to calculate this is:
Heat to warm up ice = mass of ice * specific heat of ice * change in temperature
In this case, the mass of the ice is 1.3 kg, the specific heat of ice is 2090 J/kg * K, and the change in temperature is 5 K (from -5 degrees Celsius to 0 degrees Celsius).
Heat to warm up ice = 1.3 kg * 2090 J/kg * K * 5 K = 13685 J (or 1.37 * 10^4 J)
2. Calculate the remaining heat after the ice has reached its melting point at 0 degrees Celsius. Subtract the heat required to warm up the ice from the total heat added.
Remaining heat = total heat added - heat to warm up ice
Remaining heat = (7.5 * 10^5) J - (1.37 * 10^4) J = 7.36 * 10^5 J
3. Calculate the heat required to melt the ice. This can be calculated using the formula:
Heat to melt ice = mass of ice * latent heat of fusion of ice
In this case, the mass of the ice is still 1.3 kg, and the latent heat of fusion of ice is 3.34 * 10^5 J/kg.
Heat to melt ice = 1.3 kg * (3.34 * 10^5 J/kg) = 4.34 * 10^5 J
4. Subtract the heat required to melt the ice from the remaining heat to find out how much heat is left to raise the temperature of the water.
Remaining heat = remaining heat - heat to melt ice
Remaining heat = 7.36 * 10^5 J - 4.34 * 10^5 J = 3.02 * 10^5 J
5. Finally, calculate the change in temperature of the water using the formula:
change in temperature = remaining heat / (mass of water * specific heat of water)
The mass of water can be calculated by subtracting the mass of the ice from the initial mass of the system, which is 1.3 kg.
mass of water = initial mass - mass of ice
mass of water = 1.3 kg - 1.3 kg = 0 kg
Since all the ice has melted, the mass of water is 0 kg. And the specific heat of water is 4186 J/kg * K.
change in temperature = (3.02 * 10^5 J) / (0 kg * 4186 J/kg * K) = undefined
Since the mass of the water is 0 kg, the change in temperature is undefined. This means that all the energy added to the system went into melting the ice, and the final temperature is 0 degrees Celsius (since the ice has melted into water at its melting point).