(a) Quantity of heat required to heat the lead to its melting point:
Q = m * c * (Tf - Ti)
Q = 100 kg * 120 J/kg/K * (320°C - 40°C)
Q = 3.84 x 10^7 J
(b) Additional heat required to melt the lead:
Q = m * L
Q = 100 kg * 2.5 x 10^4 J/kg
Q = 2.5 x 10^6 J
(c) Time taken to supply this additional heat:
Q = P * t
t = Q / P
t = 2.5 x 10^6 J / 10 kW
t = 250 s = 4.17 min
A block of lead of mass 100kg in cubicle and at a temperature of 40 degree Celsius was placed in a an electric furnace rate 10kw .If the melting point of lead is 320 degree Celsius .Calculate (a) the quantity of heat required to heat the lead to its melting point. (b)additional heat required to melt the lead (c)time taken to supply this additional heat (specific heat capacity of lead is 120j/kg/k, specific latent heat of fusion of lead =2.5×10^4j/k
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