Question

30. A 0.035-kg ice cube at -30.0 C is placed in 0.33 kg of 35.0 C water in a very well-insulated container. The latent heat of fusion for water is Lf=79.8 kcal/kg.

What is the final temperature of the water, in degrees Celsius?

Answer #1

first, we need to take ice from -30 C to 0 C then melt it to water

so,

Q = mcT + mLf

where, m is mass of ice .

this must be equal to heat lost by water

mcT + mLf = -M * c * Tw

mcT + mLf = -M * c * (Tf - Ti)

put the values now, we get

0.035 * 2090 * 30 + 0.035 * 334e3 = -0.33 * 4186 * (Tf - 35)

solve for Tf

Tf = 24.95 C

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the specific heat of water is 1.00 kcal/(kg · °C).

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I place an ice cube with a mass of 0.223 kg and a temperature of
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A (18) g ice cube at –15.0°C is placed in (126) g of water at
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container and assume this is in a calorimeter, i.e. the system is
thermally insulated from the surroundings. Give your answer in
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thermally insulated from the surroundings. Give your answer in
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the heat capacity of the container and assume this is in a
calorimeter, i.e. the system is thermally insulated from the
surroundings. Give your answer in oC with 3 significant
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