An object with a mass of 120 g is dropped into 800 mL of water at 0^@C. If the object cools by 20 ^@C and the water warms by 3 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
Jul 23, 2016

$1 c a l {g}^{-} 1 {C}^{-} 1$

Explanation:

Given
${m}_{o} \to \text{mass of the object} = 120 g m$

${v}_{w} \to \text{volume of water} = 800 m L$

${m}_{w} \to \text{vmass of water"=v_w xx "density of water}$

$= 800 m L \times 1 \frac{g}{m L} = 800 g$

$\Delta {t}_{w} \to \text{rise of temperature of water} = {3}^{\circ} C$

$\Delta {t}_{o} \to \text{fall of temperature of the object} = {20}^{\circ} C$

${s}_{w} \to \text{specific heat of water} = 1 c a l {g}^{-} 1 {C}^{-} 1$

s_o->"specific heat of the bodyr"=?

By considering conservation of heat energy

$\text{Heat lost by the object" ="Heat gained by water}$

${m}_{o} \times {s}_{o} \times {t}_{o} = {m}_{w} \times {s}_{w} \times {t}_{w}$

${s}_{o} = \frac{{m}_{w} \times {s}_{w} \times {t}_{w}}{{m}_{o} \times \times {t}_{o}} = \frac{800 \times 1 \times 3}{120 \times 20}$

$= 1 c a l {g}^{-} 1 {C}^{-} 1$