Question

Namaskar ♥

A calorimeter has mass 100g and specific heat 0.1 kcal/ kg° C. It contains 250g of liquid at 30° C having specific heat of 0.4 kcal/kg° C. If we drop a piece of ice of mass 10g at 0° C into it, what will be the temperature of the mixture? ​

Answer 1

$\huge{\underline{\underline{\mathfrak{Answer}}}}$

➠ Understanding :

1) Net Heat Change in this Process = 0

2) ice will be converted to water, then all will come at a equilibrium. Temperature.

➠Units are in SI system.

Steps:-

1) Given,

Mass of Caloriemeter, m = 100g = 0.1kg

Specific Heat of Caloriemeter, S (C) = 0.1 units

Mass of liquid, m. = 0.25 kg

Specific Heat of Liquid, s(L) = 0.4 units

Mass of Ice, = 0.01kg

Latent heat of fusion of Ice, L(f) = 80 units

Specific heat of water, S(w) = 1 units.

2) Heat due to fusion = m L

Heat due to change in T = ms* del(T)

Let the final Equilibrium Temperature be T. Celsius.

Q(net ) = Q(1) + Q(2) + Q(3) + Q(4)

0 = 0.1 * 0.1 (T-30) + 0.25 *0.4 (T-30) + 0.01 * 80

+ 1* 0.01 (T-0)

Multiply by 100 on both sides,

=>0 = T-30 + 10(T-30) + 80 + T

=> 12 T = 250

=> T = 20.83 °C

Therefore, Final Temperature of mixture = 20.83°C

Answer 2

Answer:

Given,

Mass of Caloriemeter, m = 100g = 0.1kg

Specific Heat of Caloriemeter, S (C) = 0.1 units

Mass of liquid, m. = 0.25 kg

Specific Heat of Liquid, s(L) = 0.4 units

Mass of Ice, = 0.01kg

Latent heat of fusion of Ice, L(f) = 80 units

Specific heat of water, S(w) = 1 units.

2) Heat due to fusion = m L

Heat due to change in T = ms* del(T)

Let the final Equilibrium Temperature be T. Celsius.

Q(net ) = Q(1) + Q(2) + Q(3) + Q(4)

0 = 0.1 * 0.1 (T-30) + 0.25 *0.4 (T-30) + 0.01 * 80

+ 1* 0.01 (T-0)

Multiply by 100 on both sides,

=>0 = T-30 + 10(T-30) + 80 + T

=> 12 T = 250

=> T = 20.83 °C