Answer:
31 grams
Explanation:
The amount of heat required to raise a mass (m, in kg) of ice from -14 °C to 10 °C is ...
heat to raise temperature of ice to 0 °C:
(2100 J/kg/K)(m kg)(14 K) = (29400m) J
heat to melt ice:
(336000 J/kg)(m kg) = (336000m) J
heat to raise melted ice to 10 °C
(4200 J/kg/K)(m kg)(10 K) = (42000m) J
So, the total heat required to raise m kg of ice from -14 °C to +10 °C is ...
(29,400 + 336,000 + 42,000)m J = (407400m) J
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The heat required to cool 200 cc = 200 g = 0.2 kg of drink by 15 K is ...
(4200 J/kg/K)(0.20 kg)(15 K) = 12600J
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The minimum amount of ice will be used if it all melts, so the two quantities of heat are the same:
407400m = 12600 . . . . joules
m = 12600/407400 ≈ 0.030928 . . . . kg
About 31 grams of ice are needed.
