Question

In: Chemistry

A calorimeter consists of a massless beaker that contains 50.0 grams of solid water (i.e., ice)...

A calorimeter consists of a massless beaker that contains 50.0 grams of solid water (i.e., ice) at −30 °C. Then 800 grams of lead at 70.0 °C are added to the calorimeter. [The specific heat of lead is 130 J kg⋅Cº. ]

a) Find the final temperature of the calorimeter after equilibrium is reached.

b) If the final temperature is 0 °C, what fraction of the ice is melted?

Expert Solution

the final temperature is zero.

According to the heat and first law of thermodynamic When thermal equilibrium is reached, the temperature of equilibrium always the same temperature of melting point of ice cubes. At that time in that mixture, some ice remains as it is.

Heat required to change temperature of 50g = 0.05 ice is determined by the specific heat of the ice

Q₁= m_iceC_ice t

The specific heat of ice is 2060 J/Kg C and initial temperature is (-30) and final temperature is (0) Then the charge of temperature is 30.

Q₁= m_iceC_ice t

=0.05 2060 30

= 3090 J

Heat required to change temperature of 800g = 0.80 lead is determined by the specific heat of the lead

The specific heat of lead is 130 J/Kg C and initial temperature is (70) and final temperature is (0) Then the charge of temperature is -70.

Q₁= m_leadC_lead t

=0.80 130 (-70)

= -7280 J

heat required to melt ice is determine by specific latent heat fusion.

The specific latent heat fusion is 334000

For ice Q₂ = m_iceL_fusion

=0.05 334000

=16700J

For lead Q₂ = m_leadL_fusion

=0.80 334000

=267200J

Specific heat of water is 4186 J/KgC

Q₃= m_iceC_water t

=0.054186T_f

₌209.3 T_f

queen_honey_blossom answered 1 month ago

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