An insulated beaker with negligible mass contains liquid water with a mass of 0.

Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.315kg and a temperature of 71.8 celsius. How much ice at a temperature of -10.7 celsius must be dropped into the water so that the final temperature of the system will be 35.0 celcius ? Take the specific heat of liquid water to be 4190J/Kg*k the specific heat of ice to be 2100 J/Kg*k , and the heat of fusion for water to be 3.34 An insulated beaker with negligible mass contains liquid water with a mass of 0.315kg and a temperature of 71.8 celsius. How much ice at a temperature of -10.7 celsius must be dropped into the water so that the final temperature of the system will be 35.0 celcius ? Take the specific heat of liquid water to be 4190J/Kg*k the specific heat of ice to be 2100 J/Kg*k , and the heat of fusion for water to be 3.34 How much ice at a temperature of -10.7 celsius must be dropped into the water so that the final temperature of the system will be 35.0 celcius ? Take the specific heat of liquid water to be 4190J/Kg*k the specific heat of ice to be 2100 J/Kg*k , and the heat of fusion for water to be 3.34

Explanation / Answer

0.315 kg * 4190 J / kg K * 36.8 = 35.8 kJ

needs to be absorbed by the ice.

1 kg of ice from -10.7 C to 35 C water will absorb =

10.7 * 2.1 kJ + 334 kJ + 35 * 4.19 kJ = 503.12 kJ/kg

35.8 kJ / 503.12 kJ / kg = 0.0711 kg of ice

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