For a 5 gm substance the heat supplied -in cals- and temperature -in -circ-C- curve is given below- Then the specific heat of substance liquid state will be -1- cal-gm-circ-C0-2- cal-gm-circ-C2-5- cal-gm-circ-CCan-t be obtain

Question

Toppr Admin · Accepted Answer

Specific heat- 100cal-5g-100oC-0-2cal-goCEnergy- 220-120- 100 cal-x3B4-T-100-x2212-0-100oCm- 5g

For a 5 gm substance the heat supplied (in cals.) and temperature (in $^{\circ} C$ ) curve is given below. Then the specific heat of substance for liquid state will be :

$1 c a l / g m^{\circ} C$
$0.2 c a l / (g m^{\circ} C$
$2.5 c a l / g m^{\circ} C$
Can't be obtain

Specific heat= $\frac{100 c a l}{(5 g) (100^{o} C)} = 0.2 c a l / g^{o} C$
Energy= 220-120= 100 cal
$δ (T) = 100 - 0 = 100^{o} C$
m= 5g

For a 5 gm substance the heat supplied (in cals.) and temperature (in ∘C) curve is given below. Then the specific heat of substance for liquid state will be :1cal/gm∘C0.2cal/(gm∘C2.5cal/gm∘CCan't be obtain

Specific heat= 100cal(5g)(100oC)=0.2cal/goCEnergy= 220-120= 100 calδ(T)=100−0=100oCm= 5g

For a 5 gm substance the heat supplied (in cals.) and temperature (in $^{\circ} C$ ) curve is given below. Then the specific heat of substance for liquid state will be :

$1 c a l / g m^{\circ} C$
$0.2 c a l / (g m^{\circ} C$
$2.5 c a l / g m^{\circ} C$
Can't be obtain

Specific heat= $\frac{100 c a l}{(5 g) (100^{o} C)} = 0.2 c a l / g^{o} C$
Energy= 220-120= 100 cal
$δ (T) = 100 - 0 = 100^{o} C$
m= 5g