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

The temperature of the kiln is 535°C.

(d) is correct option.

Step-by-step 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*0.10*(T_(f)-35)=300*1.00*(35-15)`

`T_(f)-35=(300*20)/(120*0.10)`

`T_(f)=500+35`

`T_(f)=535^(\circ)C`

Hence, The temperature of the kiln is 535°C.

Answer 2

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* c_1* (T_f-T_1)=-m_2* c_2* (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* 0.10cal/g^oC* (35-T_1)^oC=-300g* 1.00cal/g^oC* (35-15)^oC`

`T_1=535^oC`

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