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A 9.0x10^-2kg ice cube at 0.0 degrees Celsius is dropped into a styrofoam cup holding 0.35 kg of water at 13 degrees Celsius.

A. Find the final temperature of the system. Assume the cup and the surroundings can be ignored.

B. Find the amount of ice (if any) remaining.

C. Find the initial temperature of the water that would be enough to just barely melt all of the ice.

Sagot :

A. The final temperature of the system is 0°C.

B. The amount of ice (if any) remaining is  0.0331 g

C. The initial temperature of the water that would be enough to just barely melt all of the ice is 20.57 °C.

What is specific heat?

The specific heat is the amount of heat required to change the temperature by 1°C. It is denoted by C.

Heat lost or gained is represented as

Q = m C ΔT

Given, Mass of ice, mice = 9x10⁻² kg, Mass of water, mw =0.35 kg, T = 13 °C

A. If ice is in excess, final water temperature will be 0°C.

B. Specific heat of water Cp = 1000 cal/kg°C

Latent heat of ice L = 80 kcal/kg

In that case, heat lost by water =Heat gain by ice

Q = mCp x dT = mL

0.35 x 1000 x 13 = m x 80 x 1000

m = 0.0569 kg of ice.

The gram of ice remaining = 0.09 - 0.0569

                                             = 0.0331 gram of ice.

Thus,  the amount of ice remaining is 0.0331 g

C. Heat required to melt 90 gram of ice,  Q  mL

Q = 90 x 80 = 7200 cal.

If the initial temperature of water needed = T,

mCp x dT = mL

350 x T = 7200

T = 20.57 °C

Thus,  the initial temperature of the water that would be enough to just barely melt all of the ice is  20.57 °C.

Learn more about specific heat.

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