
If $ 10g $ ice at $ {0^ \circ }_C $ is mixed with $ 10g $ water at $ {20^ \circ }_C $ , the final temperature will be:-
$ \left( A \right){\text{ 50}} $ $ \left( B \right){\text{ 20}} $ $ \left( C \right){\text{ 0}} $ $ \left( D \right){\text{ 10}} $ .
Answer
457.2k+ views
Hint : Since we have to calculate the final temperature so for this we have to know the latent heat of ice and specific heat of the water. And also we already know the heat loss is equal to the heat gain. So by keeping all these in mind we can solve this problem.
Complete step by step answer
The two things to consider to solve this problem will be the Latent heat of melting ice and the other one will be the Specific heat of the water.
The amount of warmth needed to soften one gram of ice at $ {0^ \circ }_C $ liquid water $ {0^ \circ }_C $ . Alternatively, the number of warmth we need to take away to convert $ 1{\text{ }}gram $ of liquid water into the ice at $ {0^ \circ }_C $ .
The specific heat is the quantity of warmth per unit mass needed to boost the temperature by $ {1^ \circ }_C $ . The precise heat of water is $ 1cal/gram{/^ \circ }_C $ and is equal to $ 4.186joule/gram{/^ \circ }_C $ that is beyond the other common substance. As a result, water plays an awfully necessary role in temperature regulation.
As we know,
The latent heat of melting ice which is $ 80cal/gram $ and
Specific heat of water which is $ 1cal/gram{/^ \circ }_C $ .
Hence according to the question, 10g of water at 20deg cel has 200 calories.
But the requirement to melt $ 10g $ of ice would be $ 800cal $ , So at best the water would melt some of the ice and the mixture would remain at $ {0^ \circ }_C $ .
Hence we can say from the above observation that the final temperature will be $ {0^ \circ }_C $
That is Option (C).
Note
Specific heat is equal to the energy needed to vary a unit mass of fabric by $ {1^ \circ }_C $ .
Units: Energy per unit mass per degree.
Latent heat is equal to the energy needed to vary the state (gas, liquid, solid) of a unit mass of fabric.
Units: Energy per unit mass.
Both heat energy and heat of transformation are the properties of a given material. In alternative words, on every occasion the fabric is heated/cooled, regardless of however quickly or by what heating method, a similar quantity of warmth is transferred to succeed in a similar state.
Complete step by step answer
The two things to consider to solve this problem will be the Latent heat of melting ice and the other one will be the Specific heat of the water.
The amount of warmth needed to soften one gram of ice at $ {0^ \circ }_C $ liquid water $ {0^ \circ }_C $ . Alternatively, the number of warmth we need to take away to convert $ 1{\text{ }}gram $ of liquid water into the ice at $ {0^ \circ }_C $ .
The specific heat is the quantity of warmth per unit mass needed to boost the temperature by $ {1^ \circ }_C $ . The precise heat of water is $ 1cal/gram{/^ \circ }_C $ and is equal to $ 4.186joule/gram{/^ \circ }_C $ that is beyond the other common substance. As a result, water plays an awfully necessary role in temperature regulation.
As we know,
The latent heat of melting ice which is $ 80cal/gram $ and
Specific heat of water which is $ 1cal/gram{/^ \circ }_C $ .
Hence according to the question, 10g of water at 20deg cel has 200 calories.
But the requirement to melt $ 10g $ of ice would be $ 800cal $ , So at best the water would melt some of the ice and the mixture would remain at $ {0^ \circ }_C $ .
Hence we can say from the above observation that the final temperature will be $ {0^ \circ }_C $
That is Option (C).
Note
Specific heat is equal to the energy needed to vary a unit mass of fabric by $ {1^ \circ }_C $ .
Units: Energy per unit mass per degree.
Latent heat is equal to the energy needed to vary the state (gas, liquid, solid) of a unit mass of fabric.
Units: Energy per unit mass.
Both heat energy and heat of transformation are the properties of a given material. In alternative words, on every occasion the fabric is heated/cooled, regardless of however quickly or by what heating method, a similar quantity of warmth is transferred to succeed in a similar state.
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