Question

How much heat is necessary to change 480 g of ice at -19°C to water at 20°C? answer in:____kcal

Answer 1

`18.05\text{ }kcal`

Step-by-step explanation

The amount of heat necessary to change the ice to water is given by:

`Q=m(c_(ice)\Delta T_(ice)+L+c_(water)\Delta T_(water))`

where m = mass of ice = 480 g = 0.48 kg

c(ice) = specific heat capacity of ice = 2108 J/kg/K

ΔTice = change in temperature of ice = 0 - (-19) = 19 K or 19 °C

L = latent heat of fusion of ice = 33600 J/k

c(water) = specific heat capacity of water = 4186 J/kg/K

ΔTwater = change in temperature of water = 20 - 0 = 20 K or 20 °C

Therefore, the heat necessary is:

`\begin{gathered} Q=0.48([2108*19]+33600+[4186*20]) \\ \\ Q=0.48(40052+33600+83720)=0.48*157372 \\ \\ Q=75538.56\text{ }J \end{gathered}`

Convert this to kcal:

`\begin{gathered} 1\text{ }J=(1)/(4184)\text{ }kcal \\ \\ 75538.56\text{ }J=(75538.56)/(4184)\text{ }kCal=18.05\text{ }kcal \end{gathered}`

That is the answer.