Question

How much heat is necessary to change 300 g of ice at -11°C to water at 20°C? kcal

Answer 1

30 KCal

Step-by-step explanation

constant we need

`\begin{gathered} Heat\text{ capacity of water =4.186 }(J)/(g*k) \\ heat\text{ of fusion of iceUHi\rparen=333.55 J/g} \end{gathered}`

so

Step 1

To convert 300g ice -10° into ice at 0°

`\Delta H1=m*Hi=300g*=333.55(J)/(g)=100065\text{ Joules}`

Step 2

to raise the temperature to 20° C

we need to use the fomrula

letQ = m•C•ΔT.

where m is the mass C is the

`\begin{gathered} Q=m*C*\Delta T \\ Q=\text{ unknonw} \\ Q=300*4.186\text{ }(J)/(g)*(20\text{ \degree C\rparen} \\ Q=25116 \end{gathered}`

hence

the total heat need is

`\begin{gathered} Q\text{ total = 100065 Joules +25116 Joules} \\ Q\text{ totall =125181 Joules} \end{gathered}`

finally, to convert from Joules to Kcal , we need to multiply by 0.000239006

so

`\begin{gathered} \text{ Q total = 125181 J\lparen}\frac{0.000239006}{1\text{ Joules}}\text{\rparen} \\ Q\text{ total=29.91} \\ rounded \\ Q=\text{ 30 Kcal} \end{gathered}`

therefore, the answer is

30 KCal

I hope this helps you