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10-1.Thermometry, Thermal Expansion and Calorimetry
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$50\, gm$ of copper is heated to increase its temperature by $10^oC$. If the same quantity of heat is given to $10\; gm$ of water, the rise in its temperature is ........ $^oC$ (Specific heat of copper $= 420 \;Joule-kg^{-1} {°C^{-1}}$)
A
$5$
B
$6$
C
$7$
D
$8$
Solution
(a) Same amount of heat is supplied to copper and water so
${m_c}{c_c}\Delta {\theta _c} = {m_W}{c_W}\Delta {\theta _{\,W}}$
==> $\Delta {\theta _W} = \frac{{{m_c}{c_c}{{(\Delta \theta )}_c}}}{{{m_W}{c_W}}} = \frac{{50 \times {{10}^{ – 3}} \times 420 \times 10}}{{10 \times {{10}^{ – 3}} \times 4200}} = 5^\circ C$
Standard 11
Physics
