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
(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
GWyeyeye
To solve this problem, we need to calculate the heat required to heat the lead to its melting point, the additional heat required to melt the lead, and the time taken to supply this additional heat.
(a) Quantity of heat required to heat the lead to its melting point:
We can use the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
Mass of lead (m) = 100 kg
Specific heat capacity of lead (c) = 120 J/kg/K
Initial temperature (T1) = 40°C = 40 + 273 = 313 K
Melting point temperature (T2) = 320°C = 320 + 273 = 593 K
ΔT = T2 - T1 = 593 K - 313 K = 280 K
Q = mcΔT
Q = 100 kg × 120 J/kg/K × 280 K
Q = 3,360,000 J
Therefore, the quantity of heat required to heat the lead to its melting point is 3,360,000 J.
(b) Additional heat required to melt the lead:
The formula for finding the heat required to change the state of a substance is Q = mL, where Q is the heat energy, m is the mass, and L is the specific latent heat of fusion.
Given:
Specific latent heat of fusion of lead (L) = 2.5 × 10^4 J/kg
Q = mL
Q = 100 kg × 2.5 × 10^4 J/kg
Q = 2,500,000 J
Therefore, the additional heat required to melt the lead is 2,500,000 J.
(c) Time taken to supply this additional heat:
To calculate the time taken, we can use the formula t = Q/P, where t is the time, Q is the heat energy, and P is the power.
Given:
Power of the electric furnace (P) = 10 kW = 10,000 W
t = Q/P
t = (3,360,000 J + 2,500,000 J) / 10,000 W
t = 585 seconds
Therefore, the time taken to supply the additional heat and melt the lead is 585 seconds.
To calculate the quantity of heat required to heat the lead to its melting point, you can use the formula:
Q = mcΔT
Where:
Q = quantity of heat (in joules)
m = mass of the lead (in kilograms)
c = specific heat capacity of lead (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
Given:
Mass of lead, m = 100 kg
Specific heat capacity of lead, c = 120 J/kg/°C
Initial temperature, T1 = 40 °C
Final temperature, T2 = 320 °C
(a) The quantity of heat required to heat the lead to its melting point:
ΔT = T2 - T1
ΔT = 320 °C - 40 °C
ΔT = 280 °C
Q1 = mcΔT
Q1 = 100 kg * 120 J/kg/°C * 280 °C
Q1 = 3360000 J
Therefore, the quantity of heat required to heat the lead to its melting point is 3,360,000 joules.
To calculate the additional heat required to melt the lead, you can use the formula:
Q2 = mL
Where:
Q2 = additional heat required (in joules)
m = mass of the lead (in kilograms)
L = specific latent heat of fusion of lead (in joules per kilogram)
Given:
Mass of lead, m = 100 kg
Specific latent heat of fusion of lead, L = 2.5 * 10^4 J/kg
(b) The additional heat required to melt the lead:
Q2 = mL
Q2 = 100 kg * (2.5 * 10^4 J/kg)
Q2 = 2,500,000 J
Therefore, the additional heat required to melt the lead is 2,500,000 joules.
To calculate the time taken to supply this additional heat, you can use the formula:
t = Q2 / P
Where:
t = time taken (in seconds)
Q2 = additional heat required (in joules)
P = power of the electric furnace (in watts)
Given:
Additional heat required, Q2 = 2,500,000 J
Power of the electric furnace, P = 10,000 W (10 kW)
(c) The time taken to supply this additional heat:
t = Q2 / P
t = 2,500,000 J / 10,000 W
t = 250 seconds
Therefore, the time taken to supply this additional heat is 250 seconds.