Question

2.5 kg of water is taken in a calorimeter at 20°C whose mass along with stirrer is 3 kg,
600 gm of hot metal pieces at 140°C are released in to calorimeter and the final temperature
is observed as 30°C. Find the specific heat of the metal.​

Answer 1

Explanation:

${m}_{c} = \green{mass \: of \: calorimeter}$

${m}_{w} = \green{mass \: of \: water }$

${m}_{m} = \green{mass \: of \: metal }$

${t}_{c} = \blue{temp. \: of \: calorimeter }$

${t}_{w} = \blue{temp. \: of \: water }$

${t}_{m} = \blue{temp. \: of \: metal \: pieces}$

${t}_{f} = \blue{final \: temp. }$

${S}_{w} = \red{Specific \: heat \: of water}$

${S}_{m} = \red {specific \: heat \: of metal}$

Given

${t}_{w} = {t}_{c}= 20 \degree c$

${t}_{m} = 140 \degree c$

${t}_{f} = 30 \degree c$

To Find

specific heat of metal pieces

Solution

When metal pieces are added in calorimeter, heat exchange takes place

AMOUNT OF HEAT GAINED = AMOUNT OF HEAT LOST

${dQ}_{lost} ={dQ}_{gained}$

$\pink{dQ = mSdT}$

${m}_{m} .{S}_{m}({t}_m-{t}_{f}) = {m}_{c} .{S}_{c}({t}_{f} -{t}_{c}) + {m}_{w}.{S}_{w}({t}_{f}-{t}_{w})$

$\boxed{{m}_{c} .{S}_{c} = water \: equivalent}$

$600 \times {S}_{m} (140-30) = 2500 (30-20)+ 2500 \times 1 \times (30-20)$

$600 \times {S}_{m} (110) = 2500 (10) \times + 2500 \times 10$

${S}_{m} (66000) = 25000+ 25000$

${S}_{m} = \dfrac{50000}{66000}$

${S}_{m} = 0.75 \dfrac{cal}{gm \: \degree C}$