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I'm trying to use this virtual laboratory https://media.pearsoncmg.com/bc/bc_0media_chem/chem_sim/calorimetry/Calor.php

and try to calculate the specific heat capacity of ice. Based on Virtual lab when I tick on "show specific heat capacity" it shows $2.06 ~ \frac{\text{J}}{\text{g C}}$.

How to get $\mathbf{2.06 \frac{J}{g ~ C}}$?

i have tried using the equation $\mathbf{m c \Delta T_\text{ice} = m c \Delta T_\text{water}}$ but got $\mathbf{16 \frac{J}{g ~ C}}$

maybe i have a little misconception about this

note : I set

  • Substance [1] = ice ($H_{2}O$) with mass $50\ \text{g}$ and initial temperature $-20\ \text{°C}$
  • Substance [2] = water ($H_{2}O$) with a mass of $100 g$ and an initial temperature of $50\ \text{°C}$.
  • the final temperature is $3.43$ in the virtual
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You did not take in account, that the ice melts, so you need most of the energy for the heat of fusion which is 334J/g for ice from 0°C ice to 0°C water

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  • $\begingroup$ thanks sir, got it 2.066444, by adding as you described mcΔT+mL+mcΔT=mcΔT I need to study this chapter more fully next time $\endgroup$
    – Amer Barkawi
    Commented Apr 16, 2022 at 15:24

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