Question

Calculate the heat required to raise the temperature of 2.0 kg of ice at -25˚C to 70.˚C?

(I keep on getting a value of `4.0 * 10^5` but the answer key says, `1.4 * 10^6 J`.

Answer 1

`q = 4.0*10^5J`

I have a feeling you're right.

Explanation:

`q = mC_sDeltaT`

`q = (2.0*10^3g) * (2.06J/(g*°C)) * 95°C`
`q = 4.0*10^5J`

`1.4*10^6J = (2.0*10^3g) * C_s * 95°C`

If the answer key is "right", then the specific heat of ice is `7.368J*g^-1*°C^-1`. That doesn't sound accurate, so perhaps the key is wrong—maybe I'm missing something that someone else can catch!

Answer 2

The heat required is `1.40 × 10^6color(white)(l) "J"`.

Explanation:

A typical heating curve of water is shown below.

www.kentchemistry.com

There are three separate heat transfers involved in this problem:

• `q_1` = heat required to warm the ice from -25 °C to 0 °C (Point A to B in the diagram)
• `q_2` = heat required to melt the ice to water at 0 °C (Point B to C)
• `q_3` = heat required to warm the water from 0 °C to 70. °C (From Point C part way to Point D)

`q = q_1 + q_2 + q_3 = mc_1ΔT_1 + mΔ_text(fus)H + mc_3ΔT_3`

where

`q_1, q_2,` and `q_3` are the heats involved in each step

`m` is the mass of the sample

`ΔTcolor(white)(m) = T_"f" -T_"i"`

`c_1color(white)(mm) = "the specific heat capacity of ice" = "2.03 J·°C"^"-1""g"^"-1"`

`c_3color(white)(mm) = "the specific heat capacity of water" = "4.179 J·°C"^"-1""g"^"-1"`

`Δ_text(fus)H = "the enthalpy of fusion of ice" = 3.33 × 10^5 color(white)(l)"J·kg"^"-1"`

`bbq_1`

`ΔT_1 = "0 °C - (-25 °C)" = "25 °C"`

`q_1 = mc_1ΔT_1 = 2.0 × 10^3 color(red)(cancel(color(black)("g"))) × 2.03 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × 25 color(red)(cancel(color(black)("°C"))) = 1.02 × 10^5color(white)(l) "J"`

`bbq_2`

`q_2 = 2.0 color(red)(cancel(color(black)("kg"))) × 3.33 × 10^5color(white)(l) "J"·color(red)(cancel(color(black)("kg"^"-1"))) = 6.66 × 10^5color(white)(l) "J"`

`bbq_3`

`ΔT = "70. °C - 0 °C" = "70. °C"`

`q_3 = mcΔT = 2.0 × 10^3 color(red)(cancel(color(black)("g"))) × 4.179 color(white)(l)"J"·color(red)(cancel(color(black)( "°C"^"-1""g"^"-1"))) × 70. color(red)(cancel(color(black)("°C"))) = 5.85 × 10^5color(white)(l) "J"`

`q = q_1 + q_2 + q_3 = (1.02 + 6.66 + 5.85) × 10^5color(white)(l) "J" = 14.0 × 10^5color(white)(l) "J"`

`= 1.40 × 10^6color(white)(l) "J"`

The process absorbs `1.40 × 10^6color(white)(l) "J"` of heat.