A blacksmith dropped a1.5kg iron bead at 300 degree Celsius into some quantity of water , if the temperature of the water rose from 15 to 18 Celsius, what is the mass of water assuming no heat is lost to the surrounding ( take specific heat of iron as 0.46JKg -1c-1, and that of water as 4.2* 103jkg-1c-1)
To solve this problem, we need to use the concept of heat transfer and the equation for heat transfer, which is:
Q = mcΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
In this case, we have an iron bead that loses heat and transfers it to the water. The heat lost by the iron bead is equal to the heat gained by the water.
Let's first calculate the heat lost by the iron bead:
Q_iron = m_iron * c_iron * ΔT_iron
Given:
m_iron = 1.5 kg (mass of the iron bead)
c_iron = 0.46 Jkg^(-1)c^(-1) (specific heat capacity of iron)
ΔT_iron = (300 - 18) °C (change in temperature of the iron bead)
Substituting the values into the equation:
Q_iron = 1.5 kg * 0.46 Jkg^(-1)c^(-1) * (300 - 18) °C
Now, since the heat lost by the iron bead is equal to the heat gained by the water, we can calculate the mass of the water:
Q_iron = Q_water
m_water * c_water * ΔT_water = Q_iron
Given:
c_water = 4.2 * 10^3 Jkg^(-1)c^(-1) (specific heat capacity of water)
ΔT_water = (18 - 15) °C (change in temperature of the water)
Substituting the values into the equation:
m_water * 4.2 * 10^3 Jkg^(-1)c^(-1) * (18 - 15) °C = 1.5 kg * 0.46 Jkg^(-1)c^(-1) * (300 - 18) °C
Now, we can solve for m_water:
m_water = (1.5 kg * 0.46 Jkg^(-1)c^(-1) * (300 - 18) °C) / (4.2 * 10^3 Jkg^(-1)c^(-1) * (18 - 15) °C)
m_water ≈ 0.128 kg
Therefore, the mass of water is approximately 0.128 kg.