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Answer : It takes less amount of heat to metal 1.0 Kg of ice.
Solution :
The process involved in this problem are :
Now we have to calculate the amount of heat released or absorbed in both processes.
<u>For process 1 :</u>
where,
= amount of heat absorbed = ?
m = mass of water or ice = 1.0 Kg
= enthalpy change for fusion =
Now put all the given values in , we get:
<u>For process 2 :</u>
where,
= amount of heat absorbed = ?
m = mass of water = 1.0 Kg
= specific heat of liquid water =
= initial temperature =
= final temperature =
Now put all the given values in , we get:
From this we conclude that, that means it takes less amount of heat to metal 1.0 Kg of ice.
Hence, the it takes less amount of heat to metal 1.0 Kg of ice.
Answer:
Option A
250 degrees Celcius
Explanation:
If 1046J of heat energy is added to water, the water will experience a rise in temperature, at a rate that is directly proportional to its specific heat capacity.
Mathematically, this can be seen as
Where C = specific heat of water = 4.184 J/g • °C.
Q = heat energy = 1046 J
Therefore, the increase in temperature that will be experienced, is for 250 degrees Celcius