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#.

2 Answers
Jul 3, 2017

Answer:

#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!

Jul 3, 2017

Answer:

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.