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QPage 47: Q2: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q2 Solution: A piece of iron of mass 100 g is kept inside a furnace for a…

A piece of iron of mass 100 g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at 20�C. The mixture attains an equilibrium temperature of 60�C. Find the temperature of the furnace. Specific heat capacity of iron =

470 \text{ J kg}^{-1} \text{ �C}^{-1}

.

Step-by-Step Verified Solution

**Answer:** The temperature of the furnace is 950�C. **Step-by-Step Explanation:** Given:

m_{iron} = 100

g = 0.1 kg;

m_w = 240

g = 0.24 kg; Water equivalent of calorimeter,

m_{eq} = 10

g = 0.01 kg; Final temp of mixture,

T_f = 60�\text{C}

; Temp of water,

T_w = 20�\text{C}

; Specific heat capacity of iron,

C_{iron} = 470

J/kg-�C;

C_w = 4184

J/kg-�C. Let the furnace temperature be

T

. Heat loss by the iron = heat gain by the water and calorimeter

m_{iron} \times C_{iron} \times (T - T_f) = (m_w + m_{eq}) \times C_w \times (T_f - T_w)

0.1 \times 470 \times (T - 60) = (0.24 + 0.01) \times 4184 \times (60 - 20)

T = 950�\text{C}

[citation:9]

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