Question

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 water. 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

Answer 1

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

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken 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,

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

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

$m_1$ = mass of copper = 120 g

$m_2$ = mass of water = 300 g

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

$T_1$ = initial temperature of copper = ?

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

Now put all the given values in the above formula, we get:

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

$T_1=535^oC$

Therefore, the temperature of the kiln was, $535^oC$

Answer 2

Answer:

The temperature of the kiln is 535°C.

(d) is correct option.

Explanation:

Given that,

Mass of block = 120 g

Weight of water = 300 g

Initial temperature = 15°C

Final temperature = 35°C

We need to calculate the temperature of the kiln

Using formula of energy

$Q_{k}=Q_{w}$

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

Put the value 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 temperature of the kiln is 535°C.