Question

Suppose you wanted to heat a mug of water (250 mL) from room temperature 25 C to 100 C to make a cup of tea. How much energy (in units of calories and joules) would you need from your microwave?

Answer 1

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* Volume=0.99g/ml* 250ml=247.5g`

Now we have to calculate the amount of heat energy needed.

Formula used :

`Q=m* c* \Delta T=m* c* (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)* (1cal/g^oC)* (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 2

Answer: The amount of heat needed by the microwave is 18750 Cal or 78450 J

Step-by-step 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* 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* 1Cal/g.^oC* 75^oC\\\\q=18750Cal`

Converting the amount of heat absorbed in Joules, we use the conversion factor:

1 Cal = 4.184 J

So,
`18750Cal* (4.184J)/(1Cal)=78450J`

Hence, the amount of heat needed by the microwave is 18750 Cal or 78450 J