You have a well insulated cooler. You want to fill it with ice and a beverage wh

Question

You have a well insulated cooler. You want to fill it with ice and a beverage which consits mainly of water. You can ignore the volume and specific heat of the containers which hold the beverage. You put the cans of beverage, which are at 30 deg C, and the ice at 0 deg C into the cooler in the morning and you want to be able to take the cans out at 0 deg C at night, when all the ice has just melted. What should be the ratio of the mss of the ice in the cooler to the mass of the beverage in the cans in the cooler?

Explanation / Answer

Let M be the mass of ice
let m be the mass of beverage

You want the latent heat of fusion of M equal the thermal capacity of m between 0 and 30°C

Latent heat of Ice = 334 kJ/kg

Thermal capacity of water 4.182 kJ/kg°C

M(334) = m(4.182)(30 - 0)
M(334) = m(125.46)
M/m = 125.46 / 334
M/m = 0.3756...
M/m = 0.38

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