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Question

Initially a beaker has $$100$$g of water at temperature $$90^oC$$. Later another $$600g$$ of water at temperature $$20^oC$$ was poured into the beaker. The temperature, T of the water after mixing is?


A
20oC
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B
30oC
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C
45oC
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D
55oC
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E
90oC
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Solution

The correct option is C $$30^oC$$
Given,
Mass of water at $$90^oC=100$$gm
$$=100\times 10^{-3}$$kg
Mass of water at $$20^oC=600$$gm
$$=600\times 10^{-3}$$kg
From calorimetery
$$m_1s_1t_1+m_2s_2t_2=(m_1+m_2)s.T$$
$$\because s_1t_1+s_2t_2=st$$
[where, T is temperature of mixture].
$$100\times 10^{-3}\times 1\times 90+600\times 10^{-3}\times 1\times 20$$
$$=(100+600)\times 10^{-3}\times 1\times T$$
$$T=\dfrac{100\times 10^{-3}\times 90+600\times 10^{-3}\times 20}{700\times 10^{-3}}$$
$$=\dfrac{(900+12000)\times 10^{-3}}{700\times 10^{-3}}$$
$$=\dfrac{21000}{700}$$
$$=30^oC$$.

Physics

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