Math  /  Data & Statistics

Questioname: \qquad ID: A
18. A fluid that vaporizes and condenses inside the tubing of a heat pump is called the a. compressor. b. fuel. c. refrigerant. d. condenser. blem
19. How many kilojoules of heat must be transferred to a 480g480-\mathrm{g} aluminum pizza pan to raise its temperature from 22C22^{\circ} \mathrm{C} to 234C234^{\circ} \mathrm{C} ? The specific heat of aluminum in this temperature range is 0.96 J/gC0.96 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}. Show your work. 20.

As 390 g of hot milk cools in a mug, it transfers 30,000 J30,000 \mathrm{~J} of heat to the environment. What is the temperature change of the milk? The specific heat of milk is 3.9 J/gC3.9 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}. Show your work.

Studdy Solution

STEP 1

1. The specific heat formula is used to calculate the heat transfer.
2. The specific heat of aluminum is 0.96J/gC0.96 \, \text{J/g} \cdot{ }^{\circ} \text{C}.
3. The specific heat of milk is 3.9J/gC3.9 \, \text{J/g} \cdot{ }^{\circ} \text{C}.

STEP 2

1. Calculate the heat required for the aluminum pizza pan.
2. Calculate the temperature change of the milk.

STEP 3

Identify the given values for the aluminum pizza pan: - Mass (mm) = 480 g - Initial temperature (TiT_i) = 22C22^{\circ} \text{C} - Final temperature (TfT_f) = 234C234^{\circ} \text{C} - Specific heat (cc) = 0.96J/gC0.96 \, \text{J/g} \cdot{ }^{\circ} \text{C}

STEP 4

Calculate the temperature change (ΔT\Delta T) for the aluminum pizza pan:
ΔT=TfTi=234C22C=212C\Delta T = T_f - T_i = 234^{\circ} \text{C} - 22^{\circ} \text{C} = 212^{\circ} \text{C}

STEP 5

Use the formula for heat transfer:
Q=mcΔTQ = m \cdot c \cdot \Delta T
Substitute the known values:
Q=480g0.96J/gC212CQ = 480 \, \text{g} \cdot 0.96 \, \text{J/g} \cdot{ }^{\circ} \text{C} \cdot 212^{\circ} \text{C}
Calculate QQ:
Q=97,612.8J=97.6128kJQ = 97,612.8 \, \text{J} = 97.6128 \, \text{kJ}

STEP 6

Identify the given values for the milk: - Mass (mm) = 390 g - Heat transferred (QQ) = 30,000 J - Specific heat (cc) = 3.9J/gC3.9 \, \text{J/g} \cdot{ }^{\circ} \text{C}

STEP 7

Use the formula for heat transfer to find the temperature change (ΔT\Delta T):
Q=mcΔTQ = m \cdot c \cdot \Delta T
Rearrange to solve for ΔT\Delta T:
ΔT=Qmc\Delta T = \frac{Q}{m \cdot c}
Substitute the known values:
ΔT=30,000J390g3.9J/gC\Delta T = \frac{30,000 \, \text{J}}{390 \, \text{g} \cdot 3.9 \, \text{J/g} \cdot{ }^{\circ} \text{C}}
Calculate ΔT\Delta T:
ΔT19.87C\Delta T \approx 19.87^{\circ} \text{C}
The heat required for the aluminum pizza pan is 97.6128kJ97.6128 \, \text{kJ}, and the temperature change of the milk is approximately 19.87C19.87^{\circ} \text{C}.

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