Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: specific heat (ice) = 2.1 J/g·°C; specific heat (water) = 4.18 J/g·°C; ΔH fus = 6.0 kJ/mol.

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 *20 C and solving: Q=420 J

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*= 3.333 kJ= 3,333 J (being kJ=1,000 J)

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

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