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Pepsi [2]
3 years ago
8

a wooden block has a mass of 1.2 kg, a specific heat of 710, and is at a temperature of 25*C. what is the block's final temperat

ure if it's thermal energy increases by 2,130J?
Physics
1 answer:
mash [69]3 years ago
8 0

The final temperature of the block is 27.5^{\circ} C

Explanation:

The amount of thermal energy Q supplied to a substance is related to the increase in temperature of the substance, \Delta T, according to the equation

Q=mC_s \Delta T

where:

m is the mass of the substance

C_s is the specific heat capacity of the substance

In this problem, we have:

m = 1.2 kg is the mass of the block

Q = 2,130 J is the amount of energy supplied to the block

C_s = 710 J/kg^{\circ}C is the specific heat capacity of the block

Solving for \Delta T, we find the increase in temperature:

\Delta T = \frac{Q}{m C_s}=\frac{2130}{(1.2)(710)}=2.5^{\circ}C

And since the initial temperature was

T_i = 25^{\circ}C

The final temperature will be

T_f = T_i + \Delta T = 25+2.5=27.5^{\circ} C

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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Answer:

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a.) If Michael is traveling at 29 m/s, what is the resulting beat frequency that the radar gun detects?

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Approximately how much air is in a column 1-cm2 in cross section that extends from sea level to the top of the atmosphere?
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Compare the catching of two different water balloons.
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Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

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Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

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Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

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For Case B  F_{average} = 598.2×8/Δt =  4785.6/Δt

Where Δt is the same for Case A and Case B,  F_{average}  for Case B >>  F_{average}  for Case B

Therefore, Case B involves the greatest force.

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