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2 months ago

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A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa

ter. The water temperature rises from 15°C to 35°C.
Given cCu = 0.10 cal/g⋅°C, and cwater = 1.00 cal/g⋅°C, what was the temperature of the kiln?
a. 500°Cb. 360°Cc. 720°Cd. 535°C

2 answers:

ValentinkaMS [3.4K]2 months ago

7 0

Answer: The correct choice is (d) $535^oC$

Explanation:

This analysis assumes that the heat supplied by the hot body matches the heat absorbed by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = the specific heat capacity of copper = $0.10cal/g^oC$

$c_2$ = the specific heat capacity of water = $1.00cal/g^oC$

$m_1$ = the mass of copper = 120 g

$m_2$ = the mass of water = 300 g

$T_f$ = the final temperature of the mixture = $35^oC$

$T_1$ = the initial temperature of copper =?

$T_2$ = the initial temperature of water = $15^oC$

Using the provided values in the equation yields:

$120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC$

$T_1=535^oC$

Thus, the temperature of the kiln is approximately $535^oC$

Yuliya22 [3.3K]2 months ago

6 0

Answer:

The kiln temperature is 535°C.

(d) is the correct selection.

Explanation:

Provided are the following details:

Display of block mass = 120 g

Water weight = 300 g

Starting temperature = 15°C

Final temperature = 35°C

The objective is to determine the kiln temperature

Applying the energy formula

$Q_{k}=Q_{w}$

$m_{k}c_{k}\Delta T=m_{w}c_{w}\Delta T$

Substituting the given values into the formula

$120\times0.10\times(T_{f}-35)=300\times1.00\times(35-15)$

$T_{f}-35=\dfrac{300\times20}{120\times0.10}$

$T_{f}=500+35$

$T_{f}=535^{\circ}C$

Hence, the kiln temperature is calculated to be 535°C.

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