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Sagot :
Given:
The mass of ice is m = 50 g
The initial temperature of ice is
[tex]T_i=\text{ 0 }^{\circ}\text{ C}[/tex]The final temperature of the water is
[tex]T_f=\text{ 35}^{\circ}C[/tex]Explanation:
Heat is first required to convert ice to water at 0 degrees Celsius.
The heat required to convert ice to water at 0 degrees Celsius can be calculated by the formula
[tex]Q_1\text{ = mL}_f[/tex]Here, the latent heat of fusion is
[tex]L_f=\text{ 80 cal/g}[/tex]On substituting the values, the heat will be
[tex]\begin{gathered} Q_1=50\text{ }g\text{ }\times80\text{ cal/g} \\ =4000\text{ cal} \end{gathered}[/tex]Now, the heat required to increase the temperature of water from 0 degrees Celsius to 35 degrees Celsius can be calculated by the formula
[tex]\begin{gathered} Q_2=mc(\Delta\text{ T}) \\ =mc(T_f-T_i) \end{gathered}[/tex]Here, c is the specific heat of water at 1 cal/g deg C
On substituting the values, the heat required to increase the temperature will be
[tex]\begin{gathered} Q_2=50\text{ g}\times1\text{ cal/g }^{\circ}C\times(35-0) \\ =\text{ 1750 cal} \end{gathered}[/tex]The total heat will be
[tex]\begin{gathered} Q=Q_1+Q_2 \\ =4000+1750 \\ =5750\text{ cal} \end{gathered}[/tex]Final Answer:
The heat required to convert 50 grams of ice to water from 0 degrees Celsius to 35 degrees Celsius is 5750 cal
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