Question

Calculate the total energy (in kJ) absorbed when 50.5 g of ice at -15.0°C is converted into liquid water at 65.0 °C.

Answer 1

The total energy absorbed is 32.171 kilojoules.

Step-by-step explanation:

The total energy absorbed by the ice is the sum of the sensible heat of ice and water and the latent heat of fusion of the water, that is:

`Q = m\cdot [c_(i)\cdot (T_(2)-T_(1))+L_(f) + c_(w)\cdot (T_(3)-T_(2))]` (1)

Where:

`m` - Mass of the ice, in kilograms.

`c_(i)` - Specific heat of ice, in kilojoules per kilogram-degree Celsius.

`c_(w)` - Specific heat of water, in kilojoules per kilogram-degree Celsius.

`L_(f)` - Latent heat of fusion, in kilojoules per degree Celsius.

`T_(1)` - Initial temperature of water, in degrees Celsius.

`T_(2)` - Fusion point of water, in degrees Celsius.

`T_(3)` - Final temperature of water, in degree Celsius.

`Q` - Total energy absorbed, in kilojoules.

If we know that
`m = 50.5* 10^(-3)\,kg`,
`c_(i) = 2.090 \,(kJ)/(kg\cdot^(\circ)C)`,
`c_(w) = 4.180\,(J)/(kg\cdot ^(\circ)C)`,
`L_(f) = 334\,(kJ)/(kg)`,
`T_(1) = -15\,^(\circ)C`,
`T_(2) = 0\,^(\circ)C` and
`T_(3) = 65\,^(\circ)C`, then the total energy absorbed is:

`Q= (50.5* 10^(-3)\,kg)\cdot \left[\left(2.090\,(kJ)/(kg\cdot ^(\circ)C) \right)\cdot (15\,^(\circ)C) + 334\,(kJ)/(kg)+ \left(4.180\,(kJ)/(kg\cdot ^(\circ)C) \right)\cdot (65\,^(\circ)C)\right]`
`Q = 32.171\,kJ`

The total energy absorbed is 32.171 kilojoules.