An iron block of mass $1 k g$ is heated to $300^{\circ} C$ . If it is immersed in $1 k g$ of water at $30^{\circ} C$ , then what will be the change in temperature of water (in $^{\circ} C)$ ?
$[$ specific heat capacity of iron : $460 J k g^{- 1} K^{- 1}$ , water: $4200 J k g^{- 1} K^{- 1}]$

29^{\circ} C

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26.65^{\circ} C

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243.35^{\circ} C

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23.65^{\circ} C

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Solution

The correct option is B $26.65^{\circ} C$
Given,
Mass of iron block, $m_{1} = 1 k g$
Mass of water, $m_{2} = 1 k g$
Initial temperature of iron block, $T_{1, 0} = 300^{\circ} C$
Initial temperature of water, $T_{2, 0} = 30^{\circ} C$
Specific heat of iron, $c_{1} = 460 J k g^{- 1} K^{- 1}$
Specific heat of water, $c_{2} = 4200 J k g^{- 1} K^{- 1}$

Let the final temperature of the mixture be $T$ .

Heat gained by water equals the heat lost by iron sphere.
$m_{1} c_{1} (T - T_{1, 0}) = m_{2} c_{2} (T_{2, 0} - T)$
$T = \frac{m_{1} c_{1} T_{1, 0} + m_{2} c_{2} T_{2, 0}}{m_{1} c_{1} + m_{2} c_{2}}$
$T = \frac{1 \times 460 \times 300 + 1 \times 4200 \times 30}{1 \times 460 + 1 \times 4200}^{\circ} C$
$T = 56.65^{\circ} C$

Increase in the temperature of water is:
$Δ T_{2} = T - T_{2, 0}$
$Δ T_{2} = (56.65 - 30)^{\circ} C = 26.65^{\circ} C$

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An iron block of mass 1kg is heated to 300∘C. If it is immersed in 1kg of water at 30∘C, then what will be the change in temperature of water (in ∘C)? [specific heat capacity of iron : 460Jkg−1K−1, water: 4200Jkg−1K−1]

An iron block of mass $1 k g$ is heated to $300^{\circ} C$ . If it is immersed in $1 k g$ of water at $30^{\circ} C$ , then what will be the change in temperature of water (in $^{\circ} C)$ ?
$[$ specific heat capacity of iron : $460 J k g^{- 1} K^{- 1}$ , water: $4200 J k g^{- 1} K^{- 1}]$