What mass of the metal must be heated to 95 degrees celsius in order to be used to warm 2.548 L of water from 18.1 to 22.5 celsius
You didn't give the specific heat of the metal but see this post and response.
http://www.jiskha.com/display.cgi?id=1415735118
To solve this question, we need to use the heat transfer equation, which is given by:
q = m * c * ΔT
Where:
q is the amount of heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we want to find the mass of the metal, so we rearrange the equation to solve for m:
m = q / (c * ΔT)
First, let's calculate the amount of heat transferred (q) using the equation:
q = m_water * c_water * ΔT_water
Where:
m_water is the mass of water
c_water is the specific heat capacity of water (4.18 J/g°C)
ΔT_water is the change in temperature of water
Given:
m_water = 2.548 L = 2548 g (since 1 mL of water is equal to 1 g)
c_water = 4.18 J/g°C
ΔT_water = 22.5°C - 18.1°C = 4.4°C
Plugging in the values:
q = 2548 g * 4.18 J/g°C * 4.4°C
Next, we need to calculate the specific heat capacity of the metal (c_metal). This value depends on the type of metal being used. Let's assume it is 0.4 J/g°C for now.
Given:
c_metal = 0.4 J/g°C
ΔT_metal = 95°C - 22.5°C = 72.5°C
Finally, we can calculate the mass of the metal (m) using the equation:
m = q / (c_metal * ΔT_metal)
Plugging in the values calculated earlier:
m = (2548 g * 4.18 J/g°C * 4.4°C) / (0.4 J/g°C * 72.5°C)
Simplifying:
m = 24984.896 g / 29.0 J
m ≈ 861.91 g
Therefore, you would need to heat approximately 861.91 grams of the metal to 95 degrees Celsius to warm 2.548 L of water from 18.1°C to 22.5°C.