A 200-g metal container, insulated on the outside, holds 100 g of water in therm

Question

A 200-g metal container, insulated on the outside, holds 100 g of water in thermal equilibrium at 22.00degreeC. A 21-g ice cube, at the melting point, is dropped into the water, and when thermal equilibrium is reached the temperature is 15.00degreeC. Assume there is no heat exchange with the surroundings. For water, the specific heat is 4190 J kg middot K and the heat of fusion is 3.34 times 10^5 J/kg. The specific heat for the metal is closest t 2730 J/kg middot K. 4450 J/kg middot K. 3850 J/kg middot K. 4950 J/kg middot K. 5450 J/kg middot K.

Explanation / Answer

Here,

mass of metal , m = 200 gm = 0.2 Kg

mass of water , mw = 100 gm = 0.100 Kg

T = 22 degree C

let the specific heat of metal is Sm

heat lost by metal + heat lost by water = heat gain by ice

0.1 * 4186 * (22 -15) + 0.2 * Sm * (22 - 15) = 0.021 * (334 *10^3 + 4186 * 15)

sovling for Sm

Sm = 3850 J/(Kg.degree C)

the correct option is C) 3850 J/(Kg.degree C)

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