Calculate the amount of heat (in kJ) that must be absorbed to convert 108 g of ice at #0^@"C"# to water at #70^@"C"#?

1 Answer
Jun 6, 2015

You'd need 67.6 kJ of heat to convert that much ice at #0^@"C"# to water at #70^@"C"#.

So, you need to go from ice at #0^@"C"#, which is still a solid, to water at #70^@C"#, which is of course a liquid. This implies that you will go through a phase change, i.e. from ice at #0^@"C"# to water at #0^@"C"#.

As a result, you're going to have to consider two heats, one needed for the phase change and the other needed to raise the temperature of water from #0# to #70^@"C"#.

The heat needed for the phase change will be

#q_1 = m * DeltaH_"fus"#, where

#m# - the mass of ice/water;
#DeltaH_"fus"# - the enthalpy of fusion.

Plug in your values to get

#q_1 = 108cancel("g") * 333"J"/cancel("g") = "35964 J"#

The heat needed to raise the temperature of the water will be

#q_2 = m * c_"water" * DeltaT#, where

#c_"water"# - the specific heat of water;
#DeltaT# - the change in temperature, defined as the difference between the initial and the final temperature of the water.

Plug in your values to get

#q_2 = 108cancel("g") * 4.18"J"/(cancel("g") ^@cancel("C")) * (70-0)^@cancel("C") = "31600.8 J"#

The total heat needed will be

#q_"total" = q_1 + q_2#

#q_"total" = 35964 + 31600.8 = "67654.8 J"#

I'll leave the answer rounded to three sig figs and expressed in kJ, so you'll get

#q_"total" = color(green)("67.6 kJ")#