Calorimetry: Real-World Scenarios (4)

easy 2 min read

Question

A 50 g ice cube at 0°C0°\text{C} is dropped into 200 g of water at 30°C30°\text{C} in an insulated cup. Find the final temperature. (Lf=80 cal/gL_f = 80\text{ cal/g}, specific heat of water =1 cal/g⋅°C= 1\text{ cal/g·°C}.)

Solution — Step by Step

Qmelt=miceLf=50×80=4000 calQ_{melt} = m_{ice} L_f = 50 \times 80 = 4000\text{ cal}

Qwater,max=mwcΔT=200×1×30=6000 calQ_{water,max} = m_w c \Delta T = 200 \times 1 \times 30 = 6000\text{ cal}

Since 6000>40006000 > 4000, all ice melts and we still have leftover heat to warm the mixture.

After melting, we have 250 g of water (50 g from ice at 0°C0°\text{C} + 200 g at some intermediate temp). Heat balance:

mwc(30Tf)=miceLf+micec(Tf0)m_w c (30 - T_f) = m_{ice} L_f + m_{ice} c (T_f - 0)

200(30Tf)=4000+50Tf200(30 - T_f) = 4000 + 50 T_f

6000200Tf=4000+50Tf    2000=250Tf6000 - 200 T_f = 4000 + 50 T_f \implies 2000 = 250 T_f

Tf=8°CT_f = 8°\text{C}

Final answer: Tf=8°CT_f = 8°\text{C}

Why This Works

Calorimetry is bookkeeping: heat lost by hot stuff = heat gained by cold stuff. The trick with phase changes is checking whether the cold body has enough mass to absorb all the available heat for melting. If not, only part melts and the answer is 0°C.

Always do the “is there enough heat to melt all of it?” check first — saves you from setting up the wrong equation.

Alternative Method

Set the final temperature symbolically and write Qlost=QgainedQ_{lost} = Q_{gained} in one line. Same answer, but the partial-melt check still has to happen first.

Common Mistake

Students forget to warm the melted ice from 0°C0°\text{C} to the final temperature. The 50 g of “ice water” needs 50×1×Tf50 \times 1 \times T_f calories — without that term, you get Tf=10°CT_f = 10°\text{C} instead of 8°C8°\text{C}.

If Qwater,max<QmeltQ_{water,max} < Q_{melt}, the answer is automatically Tf=0°CT_f = 0°\text{C} with some unmelted ice remaining. Don’t bother solving the heat-balance equation in that case.

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