# help, please?

An eager scientist on a hot summer day needs to determine how much ice to buy. She has a cooler which is filled with 32 cans of soda at a temperature of 81.7∘F. Each can has a mass of 423 g. The scientist wants the temperature of the drinks to be 41.5∘F. If there is no heat lost by the cooler and ignoring any heat lost to the soda containers, what mass 𝑚ice of ice needs must she add to her cooler? Assume the temperature of the ice is 32.0 °F.

### 1 Answer

- oubaasLv 71 month ago
Convert all temperatures to °C with the formula C = (F-32)*5/9 , therefore :

81.7 F = (81.7-32)*5/9 = 27.61 °C

41.5 F = (41.5-32)*5/9 = 5.28 °C

32 F = 0°C Hi = 3.34*10^5 joule/kg

cs = 4.186*10^3 joule /(kg*°C)

Es = 0.423*32*4186*27.61 joule

Ei = -mi*3.34*10^5 joule

Es+Ei = (ms+mi)*cs*5.28

13.536*4.186*10^3*27.61-mi*3.34*10^5 = (13.536+mi)*4.186*10^3*5.28

13.536*4.186*10^3*(27.61-5.28) = mi(*3.34*10^5+0.22*10^5)

1.265*10^6 = mi*3.56*10^5

mi = 12.65/3.56 = 3.55 kg