# 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 $4.0 \cdot {10}^{5}$ but the answer key says, $1.4 \cdot {10}^{6} J$.

Jul 3, 2017

$q = 4.0 \cdot {10}^{5} J$

I have a feeling you're right.

#### Explanation:

$q = m {C}_{s} \Delta T$

q = (2.0*10^3g) * (2.06J/(g*°C)) * 95°C
$q = 4.0 \cdot {10}^{5} J$

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!

Jul 3, 2017

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

#### Explanation:

A typical heating curve of water is shown below.

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"

${c}_{1} \textcolor{w h i t e}{m m} = \text{the specific heat capacity of ice" = "2.03 J·°C"^"-1""g"^"-1}$

${c}_{3} \textcolor{w h i t e}{m m} = \text{the specific heat capacity of water" = "4.179 J·°C"^"-1""g"^"-1}$

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

${\boldsymbol{q}}_{1}$

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"

${\boldsymbol{q}}_{2}$

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"

${\boldsymbol{q}}_{3}$

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.