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2 years ago

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.

Physics

1 answer:

Alexxandr [17]2 years ago

5 0

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.

<h3>What is specific heat?</h3>

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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