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
B
30oC
C
45oC
D
55oC
E
90oC

Solution

## The correct option is C $$30^oC$$Given,Mass of water at $$90^oC=100$$gm$$=100\times 10^{-3}$$kgMass of water at $$20^oC=600$$gm$$=600\times 10^{-3}$$kgFrom 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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