Question

How many joules are required to heat 15 grams of ice from -200⁰ to 1400⁰ C?

Answer 1

Answer: A heat energy of 50160 joules are required to heat 15 grams of ice from
`-200^(o)C` to
`1400^(o)C`.

Step-by-step explanation:

Given : mass of ice = 15 g

Convert grams into kg as follows,

`1 g = 0.001 kg\\15 g = 15 g * (0.001 kg)/(1 g)\\= 0.015 kg`

Initial temperature =
`-200^(o)C`

Final temperature =
`1400^(o)C`

Formula used to calculate the amount of heat energy is as follows.

`q = m * C * (T_(2) - T_(1))`

where,

q = heat energy

m = mass of substance

C = specific heat of substance (ice here has C =
`2090 J/kg ^(o)C`)

`T_(1)` = initial temperature

`T_(2)` = final temperature

Substitute the values into above formula as follows.

`q = m * C * (T_(2) - T_(1))\\= 0.015 kg * 2090 J/kg^(o)C * [1400 - (-200)]^(o)C\\= 50160 J`

Thus, we can conclude that 50160 joules are required to heat 15 grams of ice from
`-200^(o)C` to
`1400^(o)C`.