$400 g$ of iron is heated to increase its temperature by $10^{\circ} C$ . If the same quantity of heat is given to $200 g$ of water, the rise in its temperature is ___ $^{\circ} C$ (specific heat of iron = $420 J / k g /^{\circ} C$ , specific heat of water = $4200 J / k g /^{\circ} C$ )

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Solution

Given

Mass of iron = $400 g$

Mass of water = $200 g$

specific heat of copper = $420 J / k g /^{\circ} C$

specific heat of water = $4200 J / k g /^{\circ} C$

Change in temperature = $Δ T$

Same amount of heat is supplied to copper and water; so $Q_{1}$ = $Q_{2}$

$m_{i} C_{i} Δ T_{i}$ = $m_{w} C_{w} Δ T_{w}$

$\Rightarrow$ $(Δ T)_{w}$ = $\frac{m_{i} C_{i} Δ T_{i}}{m_{w} c_{w}}$ = $\frac{400 \times 10^{- 3} \times 420 \times 10}{200 \times 10^{- 3} \times 4200}$ = $2^{\circ} C$

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$50 g$ of copper is heated to increase its temperature by $10^{\circ} C$ . If the same quantity of heat is given to $10 g$ of water, then what would be the rise in its temperature?
(specific heat of copper $= 420 J / k g /^{\circ} C$ , specific heat of water $= 4200 J / k g /^{\circ} C$ )

$50 g$ of copper is heated to increase its temperature by $10^{\circ} C$ . If the same quantity of heat is given to $10 g$ of water, the rise in its temperature is (specific heat of copper = $420 J / k g /^{\circ} C$ , specific heat of water = $4200 J / k g /^{\circ} C$ )