Verified Answer :
Answer : The amount of heat energy needed from microwave will be, 18412.5 calories and 77037.9 joules
Solution : Given,
Initial temperature = \(25^oC\)
Final temperature = \(100^oC\)
Specific heat of water = \(1cal/g^oC\)
Density of water = 0.99 g/ml
Volume of water = 250 ml
First we have to calculate the mass of substance.
\(Mass=Density\times Volume=0.99g/ml\times 250ml=247.5g\)
Now we have to calculate the amount of heat energy needed.
Formula used :
\(Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)\)
where,
Q = heat energy gained
c = specific heat of water
\(T_1\) = initial temperature
\(T_2\) = final temperature
Now put all the given values in the above formula, we get
\(Q=(247.5g)\times (1cal/g^oC)\times (100-25)^oC\)
\(Q=18412.5cal\)
The amount of heat energy needed in calories = 18412.5 cal
The amount of heat energy needed in joules = (4.184 × 18412.5) = 77037.9 J
Conversion used : (1 calorie = 4.184 joules)
Therefore, the amount of heat energy needed from microwave will be, 18412.5 calories and 77037.9 joules
Answer: The amount of heat needed by the microwave is 18750 Cal or 78450 J
Explanation:
To calculate the mass of water, we use the equation:
\(\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}\)
Density of water = 1 g/mL
Volume of water = 250 mL
Putting values in above equation, we get:
\(1g/mL=\frac{\text{Mass of water}}{250mL}\\\\\text{Mass of water}=(1g/mL\times 250mL)=250g\)
To calculate the amount of heat absorbed, we use the equation:
\(q=mc\Delta T\)
where,
m = mass of water = 250 g
c = specific heat capacity of water = 1 Cal/g°C
\(\Delta T\) = change in temperature = \(T_2-T_1=100^oC-25^oC=75^oC\)
Putting values in above equation, we get:
\(q=250g\times 1Cal/g.^oC\times 75^oC\\\\q=18750Cal\)
Converting the amount of heat absorbed in Joules, we use the conversion factor:
1 Cal = 4.184 J
So, \(18750Cal\times \frac{4.184J}{1Cal}=78450J\)
Hence, the amount of heat needed by the microwave is 18750 Cal or 78450 J