Math / Algebra

Question41.0 kg of water per second flows through the cooling system of a diesel engine. The difference between the temperatures of the incoming and outgoing water is 5.0 K . How much energy is being removed each second?

Studdy Solution

STEP 1

1. The mass flow rate of water is 41.0 kg/s.
2. The temperature difference between incoming and outgoing water is 5.0 K.
3. The specific heat capacity of water is approximately $4,186 \, \text{J/(kg} \cdot \text{K)}$ .
4. We need to calculate the energy removed per second, which is equivalent to power in watts.

STEP 2

1. Understand the relationship between heat energy, mass, specific heat capacity, and temperature change.
2. Define the formula for calculating energy.
3. Substitute the known values into the formula.
4. Calculate the energy removed per second.

STEP 3

Understand the relationship between heat energy, mass, specific heat capacity, and temperature change. The formula to calculate the heat energy $Q$ removed is:
$Q = m \cdot c \cdot \Delta T$
where: - $Q$ is the heat energy in joules, - $m$ is the mass flow rate in kg/s, - $c$ is the specific heat capacity in J/(kg·K), - $\Delta T$ is the temperature change in K.

STEP 4

Define the formula for calculating energy. We will use the formula:
$Q = m \cdot c \cdot \Delta T$

STEP 5

Substitute the known values into the formula:
- $m = 41.0 \, \text{kg/s}$ - $c = 4,186 \, \text{J/(kg} \cdot \text{K)}$ - $\Delta T = 5.0 \, \text{K}$
$Q = 41.0 \cdot 4,186 \cdot 5.0$

STEP 6

Calculate the energy removed per second:
$Q = 41.0 \cdot 4,186 \cdot 5.0$
$Q = 858,230 \, \text{J/s}$
Thus, the energy being removed each second is:
$\boxed{858,230 \, \text{J/s}}$

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Question41.0 kg of water per second flows through the cooling system of a diesel engine. The difference between the temperatures of the incoming and outgoing water is 5.0 K . How much energy is being removed each second?

Studdy Solution

STEP 1

STEP 2

1. Understand the relationship between heat energy, mass, specific heat capacity, and temperature change.
2. Define the formula for calculating energy.
3. Substitute the known values into the formula.
4. Calculate the energy removed per second.

STEP 3

STEP 4

Define the formula for calculating energy. We will use the formula:
$Q = m \cdot c \cdot \Delta T$

STEP 5

Substitute the known values into the formula:
- $m = 41.0 \, \text{kg/s}$ - $c = 4,186 \, \text{J/(kg} \cdot \text{K)}$ - $\Delta T = 5.0 \, \text{K}$
$Q = 41.0 \cdot 4,186 \cdot 5.0$

STEP 6

Calculate the energy removed per second:
$Q = 41.0 \cdot 4,186 \cdot 5.0$
$Q = 858,230 \, \text{J/s}$
Thus, the energy being removed each second is:
$\boxed{858,230 \, \text{J/s}}$

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Question41.0 kg of water per second flows through the cooling system of a diesel engine. The difference between the temperatures of the incoming and outgoing water is 5.0 K . How much energy is being removed each second?

Studdy Solution

STEP 1

STEP 2

1. Understand the relationship between heat energy, mass, specific heat capacity, and temperature change.2. Define the formula for calculating energy.3. Substitute the known values into the formula.4. Calculate the energy removed per second.

STEP 3

STEP 4

Define the formula for calculating energy. We will use the formula:Q=m⋅c⋅ΔT Q = m \cdot c \cdot \Delta T Q=m⋅c⋅ΔT

STEP 5

Substitute the known values into the formula:- m=41.0 kg/s m = 41.0 \, \text{kg/s} m=41.0kg/s - c=4,186 J/(kg⋅K) c = 4,186 \, \text{J/(kg} \cdot \text{K)} c=4,186J/(kg⋅K) - ΔT=5.0 K \Delta T = 5.0 \, \text{K} ΔT=5.0KQ=41.0⋅4,186⋅5.0 Q = 41.0 \cdot 4,186 \cdot 5.0 Q=41.0⋅4,186⋅5.0

STEP 6

Calculate the energy removed per second:Q=41.0⋅4,186⋅5.0 Q = 41.0 \cdot 4,186 \cdot 5.0 Q=41.0⋅4,186⋅5.0Q=858,230 J/s Q = 858,230 \, \text{J/s} Q=858,230J/sThus, the energy being removed each second is:858,230 J/s \boxed{858,230 \, \text{J/s}} 858,230J/s​

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1. Understand the relationship between heat energy, mass, specific heat capacity, and temperature change.
2. Define the formula for calculating energy.
3. Substitute the known values into the formula.
4. Calculate the energy removed per second.

Define the formula for calculating energy. We will use the formula:
$Q = m \cdot c \cdot \Delta T$

Substitute the known values into the formula:
- $m = 41.0 \, \text{kg/s}$ - $c = 4,186 \, \text{J/(kg} \cdot \text{K)}$ - $\Delta T = 5.0 \, \text{K}$
$Q = 41.0 \cdot 4,186 \cdot 5.0$

Calculate the energy removed per second:
$Q = 41.0 \cdot 4,186 \cdot 5.0$
$Q = 858,230 \, \text{J/s}$
Thus, the energy being removed each second is:
$\boxed{858,230 \, \text{J/s}}$