3.A metal of mass 200g at temperature of 100°c is placed in 100g of water at 20°c in a container of negligible heat capacity.If the final steady temperature is 30°c.Calculate the specific heat capacity of the metal ( specific heat capacity of water 4200jkg-¹k-¹).
To calculate the specific heat capacity of the metal, we can use the principle of conservation of heat energy. The heat gained by the water is equal to the heat lost by the metal.
Let's first calculate the heat gained by the water using the formula:
Q1 = m1 * c1 * ΔT1
Where:
Q1 = heat gained by water
m1 = mass of water = 100g
c1 = specific heat capacity of water = 4200 J/kg°C
ΔT1 = change in temperature of water = (final temperature - initial temperature) = (30°C - 20°C) = 10°C
Plugging the values into the formula:
Q1 = 100g * 4200 J/kg°C * 10°C
Q1 = 420,000 J
Next, let's calculate the heat lost by the metal using the formula:
Q2 = m2 * c2 * ΔT2
Where:
Q2 = heat lost by metal
m2 = mass of metal = 200g
c2 = specific heat capacity of metal (unknown)
ΔT2 = change in temperature of metal = (final temperature - initial temperature) = (30°C - 100°C) = -70°C
Plugging the values into the formula:
420,000 J = 200g * c2 * -70°C
420,000 J = -14,000g°C * c2
Simplifying:
c2 = 420,000 J / (-14,000g°C)
c2 = -30 J/g°C
Therefore, the specific heat capacity of the metal is -30 J/g°C.