HOW MUCH HEAT IS ABSORBED WHEN THE BLOCK OF COPPER OF MASS 0.05kg and specific capacity 390j/kg is heated for 20 degree celsius and 70 degree celsius
heat=mass*specificheat*(Tfinal-Tinitial)
To calculate the heat absorbed by the block of copper, you can use the formula:
Q = mcΔT
Where:
Q = Heat absorbed (in joules)
m = Mass of the block of copper (in kilograms)
c = Specific heat capacity of copper (in joules per kilogram per degree Celsius)
ΔT = Change in temperature (in degrees Celsius)
Given:
m = 0.05 kg (mass of the copper block)
c = 390 J/kg°C (specific heat capacity of copper)
ΔT = 70°C - 20°C = 50°C (change in temperature)
Now, let's substitute the given values into the formula:
Q = (0.05 kg) * (390 J/kg°C) * (50°C)
Q = 975 joules
Therefore, the block of copper absorbs 975 joules of heat when heated from 20 degrees Celsius to 70 degrees Celsius.