
A copper block of mass $2.5kg$ is heated in a furnace to a temperature of ${500^ \circ }C$ and then placed on a large ice block. What is the maximum amount of ice that can melt?
(Specific heat of copper $ = 0.39J{g^{ - 1}}{K^{ - 1}}$, heat of fusion of water $ = 335J{g^{ - 1}}$)
Answer
598.5k+ views
Hint: Total energy would be conserved I.e. Total energy consumed in melting ice = total heat supplied to the body.
Formula Used:
Latent Heat of fusion = Heat energy required to transform 1 g of ice to water.
Heat transferred $Q = mC\Delta T$ where C is specific heat …… (a)
Complete step by step answer:
Given,
Latent heat of fusion $L = 335J{g^{ - 1}}$
Specific Heat of copper $C = 0.39J{g^{ - 1}}{K^{ - 1}}$
Mass of copper block $M = 2.5kg$
Temperature rise up to, ${T_f} = {500^ \circ }C$
Step 1 of 6:
Temperature of ice would be the reference temperature in our case and we know the final temperature of the copper block.
So, $\Delta T = 500K$
Step 2 of 6:
Total heat content of Copper block given by equation (a), $Q = mC\Delta T$ …… (1)
Step 3 of 6:
Putting values given in ques in equation (1) we get $Q = 2.5kg \times 0.39J{g^{ - 1}}{K^{ - 1}} \times 500K$
\[ \Rightarrow Q = 487.5 \times {10^3}J\] …… (2)
Step 4 of 6:
Using Total energy conservation of the system,
Then, total heat lost by copper = total heat gained by ice for melting …… (3)
Step 5 of 6:
Substituting equation (2) and latent heat given in question in equation (3) we get
$Q = {m_{ice}}{L_{ice}}$
$ \Rightarrow {m_{ice}} = \dfrac{Q}{L}$
$ \Rightarrow {m_{ice}} = \dfrac{{487.5 \times {{10}^3}J}}{{335J{g^{ - 1}}}} = 1455.22g = 1.455kg$
Final Answer:
Mass of melted ice =$1455.22g$
Additional Information: There are various other types of latent heats like latent heat of vaporization, fusion, condensation etc. All these latent heats come into consideration at phase transition between states of substance. These phase change processes differ from one substance to another depending upon their molecular properties.
Note: Energy conservation law in case of heat transfer assumes that there is no energy loss or leakage in other processes than heat transfer via conduction between copper to ice.
Formula Used:
Latent Heat of fusion = Heat energy required to transform 1 g of ice to water.
Heat transferred $Q = mC\Delta T$ where C is specific heat …… (a)
Complete step by step answer:
Given,
Latent heat of fusion $L = 335J{g^{ - 1}}$
Specific Heat of copper $C = 0.39J{g^{ - 1}}{K^{ - 1}}$
Mass of copper block $M = 2.5kg$
Temperature rise up to, ${T_f} = {500^ \circ }C$
Step 1 of 6:
Temperature of ice would be the reference temperature in our case and we know the final temperature of the copper block.
So, $\Delta T = 500K$
Step 2 of 6:
Total heat content of Copper block given by equation (a), $Q = mC\Delta T$ …… (1)
Step 3 of 6:
Putting values given in ques in equation (1) we get $Q = 2.5kg \times 0.39J{g^{ - 1}}{K^{ - 1}} \times 500K$
\[ \Rightarrow Q = 487.5 \times {10^3}J\] …… (2)
Step 4 of 6:
Using Total energy conservation of the system,
Then, total heat lost by copper = total heat gained by ice for melting …… (3)
Step 5 of 6:
Substituting equation (2) and latent heat given in question in equation (3) we get
$Q = {m_{ice}}{L_{ice}}$
$ \Rightarrow {m_{ice}} = \dfrac{Q}{L}$
$ \Rightarrow {m_{ice}} = \dfrac{{487.5 \times {{10}^3}J}}{{335J{g^{ - 1}}}} = 1455.22g = 1.455kg$
Final Answer:
Mass of melted ice =$1455.22g$
Additional Information: There are various other types of latent heats like latent heat of vaporization, fusion, condensation etc. All these latent heats come into consideration at phase transition between states of substance. These phase change processes differ from one substance to another depending upon their molecular properties.
Note: Energy conservation law in case of heat transfer assumes that there is no energy loss or leakage in other processes than heat transfer via conduction between copper to ice.
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