2020-04-15T07:29:31-04:00
What mass of ice at -14 Celsius will be needed to cool 200cm cube of an orange drink from 25 Celsius to 10 Celsius.( Specific fusion of ice= 3.36×10^5.specific heat capacity of ice= 2100. Specific heat capacity of water= 4200Jkg^-1
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2020-04-16T08:57:58-0400
m i ( L + c i ( T 0 − T i ) + c w ( T − T 0 ) ) = m w c w ( T w − T ) m_i(L+c_i(T_0-T_i)+c_w(T-T_0))=m_wc_w(T_w-T) m i ( L + c i ( T 0 − T i ) + c w ( T − T 0 )) = m w c w ( T w − T )
m i = ρ V w c w ( T w − T ) L + c i ( T 0 − T i ) + c w ( T − T 0 ) m_i=\rho V_w\frac{c_w(T_w-T)}{L+c_i(T_0-T_i)+c_w(T-T_0)} m i = ρ V w L + c i ( T 0 − T i ) + c w ( T − T 0 ) c w ( T w − T )
m i = ( 1000 ) ( 200 ⋅ 1 0 − 6 ) ( 4200 ) ( 25 − 10 ) 3.36 ⋅ 1 0 5 + 2100 ( 0 − ( − 14 ) ) + 4200 ( 10 − 0 ) m_i=\frac{(1000)(200\cdot10^{-6})(4200)(25-10)}{3.36\cdot10^{5}+2100(0-(-14))+4200(10-0)} m i = 3.36 ⋅ 1 0 5 + 2100 ( 0 − ( − 14 )) + 4200 ( 10 − 0 ) ( 1000 ) ( 200 ⋅ 1 0 − 6 ) ( 4200 ) ( 25 − 10 )
m i = 0.031 k g = 31 g m_i=0.031\ kg=31\ g m i = 0.031 k g = 31 g
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