5 kg of steel is immersed in water at 2 degrees Celsius. what mass of water is required if the final temperature is 40 degrees Celsius?
To determine the mass of water required to raise the temperature of 5 kg of steel from 2 degrees Celsius to 40 degrees Celsius, we need to calculate the heat transfer between the steel and water.
The heat transfer equation can be expressed as:
Q = mc∆T
Where:
Q is the heat transferred (in Joules)
m is the mass of the substance (in kg)
c is the specific heat capacity of the substance (in J/kg·°C)
∆T is the change in temperature (in °C)
For steel, the specific heat capacity (c) is approximately 490 J/kg·°C, whereas for water, the specific heat capacity is approximately 4186 J/kg·°C.
First, let's calculate the heat transferred from the steel to the water:
Q_steel = m_steel * c_steel * ∆T_steel
= 5 kg * 490 J/kg·°C * (40°C - 2°C)
= 5 kg * 490 J/kg·°C * 38°C
= 93100 J
Next, let's calculate the mass of water required to absorb this amount of heat:
Q_steel = Q_water
93100 J = m_water * c_water * ∆T_water
We can rearrange this equation to solve for the mass of water:
m_water = Q_steel / (c_water * ∆T_water)
= 93100 J / (4186 J/kg·°C * (40°C - 2°C))
= 93100 J / (4186 J/kg·°C * 38°C)
= 1.47 kg (approximately)
Therefore, approximately 1.47 kg of water is required to raise the temperature of 5 kg of steel from 2 degrees Celsius to 40 degrees Celsius.