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ICSE Class 10 Physics 2013 Paper

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Question : 56 of 70
Marks: +1, -0
A calorimeter of mass 50 g50\text{ g} and specific heat capacity 0.42 J g−1 ∘C−10.42\,\text{J}\,\text{g}^{-1}\,{}^{\circ}\text{C}^{-1} contains some mass of water at 20∘ C20^{\circ}\,\text{C}. A metal piece of mass 20 g20\text{ g} at 100∘C100^{\circ}\text{C} is dropped into the calorimeter. After stirring, the final temperature of the mixture is found to be 22∘C22^{\circ}\text{C}. Find the mass of water used in the calorimeter.
[specific heat capacity of the metal piece == 0.3 Jg−1 ∘C−10.3\,\text{Jg}^{-1}\,{}^{\circ}\text{C}^{-1}
specific heat capacity of water =4.2 J g−1=4.2\,\text{J}\,\text{g}^{-1} ∘ C−1]^{\circ}\,\text{C}^{-1}]
Solution:  
Heat energy given by metal piece
  =m⋅c⋅ΔT1\;=m\cdot c\cdot \Delta T_1
  =20×0.3×(100−22)\;=20\times 0.3\times (100-22)
  =468 Joule\;=468\ \text{Joule}
Heat energy gained by water
=mw×cw×ΔT2=m_w \times c_w \times \Delta T_2
  =mw×4.2×(22−20)\;=m_w \times 4.2 \times (22-20)
  =mw×8.4   Joule   \;=m_w \times 8.4 \;\text{ Joule }\;
Heat energy gained by calorimeter
  =mc×cc×ΔT2\;=m_c \times c_c \times \Delta T_2
  =50×0.42×(22−20)\;=50 \times 0.42 \times (22-20)
  =42   Joule   \;=42 \;\text{ Joule }\;
By principle of calorimetry
   Heat lost   =   Heat gained   \; \text{ Heat lost }\; = \; \text{ Heat gained }\;
Heat energy given by metal == Heat energy gained by water + Heat energy gained by calorimeter
468  =mw×8.4+42468\;=m_w \times 8.4+42
mw  =  468−428.4m_w\;=\;\frac{468-42}{8.4}
  =50.7 gm.\;=50.7\ \text{gm}.
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