what is the mass and volume of water that supplied with 2.26 × 10⁶ J of heat to increase its temperature by 70⁰C?
Answers
Mass is 7.71 kg
3
Volume is 0.0077 m³
Explanation:
Step 1: Enumerate the given values
Given:
Q = 2.26 * 10 ^ 6 * J
AT = 70°C
cp of water = 4.186 J/g°C
density of water = 1000 kg/m³
Step 2: Identify what is being asked.
Required:
Mass
Volume
Step 3: Solve for Mass and Volume
Equation:The formula for Heat Energy is given by:
Q = mcpAT
where: Q - Heat Energy
m - mass
cp-specific heat AT = change in temperature, T2 -
T1
Solution:
Find the mass by substituting the value of Q, cp and AT in the formula m = (2.26 x 10°J)/(4.186 J/g°C)(70°C) m = 7712.78 g or 7.71278 kg or 7.71 kg
2.26 x 100J = (m)(4.186 J/g°C)(70°C)
Find the volume using the density of water which is by using the formula for density
p = m/v
where: p - density
m- volume
V - volume
Solution:
Substitute the value of p and m in the equation:
1000 kg / (m ^ 3) = (7.71278 kg) / v
v = (7.71278 kg) / (1000 kg / (m ^ 3))
v = 0.00771278m ^ 3 or 0.0077 m³
Answer:
Mass is 7.71 kg
Volume is 0.0077 m³
Explanation:
Step 1: Enumerate the given values.
Given:
Q = 2.26 x 10⁶J
∆T = 70°C
cp of water = 4.186 J/g°C
density of water = 1000 kg/m³
Step 2: Identify what is being asked.
Required:
Mass
Volume
Step 3: Solve for Mass and Volume
Equation:
The formula for Heat Energy is given by:
Q = mcp∆T
where: Q - Heat Energy
m - mass
cp - specific heat
∆T = change in temperature, T2 - T1
Solution:
Find the mass by substituting the value of Q, cp and ∆T in the formula
2.26 x 10⁶J = (m)(4.186 J/g°C)(70°C)
m = (2.26 x 10⁶J)/(4.186 J/g°C)(70°C)
m = 7712.78 g or 7.71278 kg or 7.71 kg
Find the volume using the density of water which is by using the formula for density
ρ = m/v
where: ρ - density
m - volume
v - volume
Solution:
Substitute the value of ρ and m in the equation:
1000 kg/m³ = (7.71278 kg)/v
v = (7.71278 kg)/(1000 kg/m³)
v = 0.00771278 m³ or 0.0077 m³
#CarryOnLearning
" I hope it helps "