Question

mass of water is 0.3kg

An insulated beaker with negligible mass contains a mass of 0.300 kg of water at a temperature of 84.1 °C. - Part A How many kilograms of ice at a temperature of – 10.3 °C must be dropped in the water to make the final temperature of the system 29.9 °C? Take the specific heat for water to be 4190 J/(kg · K), the specific heat for ice to be 2100 J/(kg · K), and the heat of fusion for water to be 334 kJ/kg . OVALO M O 2 ? mice = kg Submit Request Answer

Answer 1

Let,

• $T_{water}$    : be the initial temperature of the water.
• $T_{ice}$        : be the initial temperature of the ice.
•          : be the final temperature of the system.
•           : be the latent heat of fusion of water.
• $C_{ice}$      : be the specific heat of ice.
• $C_{water}$ : be the specific heat of water.
• $m_{water}$ : mass of the water.
• $m_{ice}$      : mass of the ice.

Given:-

• $\small m_{water} = 0.3 \; kg$
• $\small T_{water} = 84.1 \: ^{\circ}C$
• $\small T_{ice} = -10.3 \: ^{\circ}C$
• $\small T = 29.9 \: ^{\circ}C$
• $\small C_{water} = 4190 \; J/\left (kg \cdot K \right )$
• $\small C_{ice} = 2100 \; J/\left (kg \cdot K \right )$
• $\small L = 334 \; kJ/kg = 334 \times 10^{3} \; J/kg$

When the system reaches the final temperature T then the total heat flow for the system becomes zero,

Where,

• $Q_{total}$ : be the total heat flow for the system.
• $Q_{ice}$     : heat for the ice.
• $Q_{water}$ : heat for water.

Since the ice will go through three stages in this process, hence the $Q_{ice}$ for ice, which has initial temperature of -10.3^oC and final temperature 0^oC, given by,

$Q_{ice} =\left ( m_{ice}\times C_{ice} \times \Delta T_{ice} \right ) + \left ( m_{ice} \times L \right ) + \left ( m_{ice} \times C_{water} \times \Delta T_{melted} \right )$

$Q_{ice} = m_{ice} \left [C_{ice} \times \Delta T_{ice} + L + C_{water} \times \Delta T_{melted} \right ]$

$Q_{ice} = m_{ice} \left [2100 \times \left ( 0^{\circ}-\left ( -10.3^{\circ} \right ) \right ) + \left (334 \times 10^{3} \right ) + 4190 \times \left ( 29.9^{\circ} -0 ^{\circ}\right ) \right ]$

$\mathbf{Q_{ice} = m_{ice} \left (480911 \right )}$

This is the heat gain by the ice.

Now, $Q_{water}$ for the water, which drops the temperature from 84.1^oC to 29.9^oC is give by,

$Q_{water} = m_{water} \times C_{water} \times \Delta T_{water}$

$Q_{water} = 0.3 \times 4190 \times \left ( 29.9^{\circ} - 84.1^{\circ} \right )$

$\mathbf{Q_{water} = -68129.4 \; J}$

The negative sign indicates that the heat is lost by the water.

Now the total heat is given by,

$m_{ice} \left (480911 \right ) -68129.4 = 0$

$m_{ice} = \frac{68129.4}{480911}$

$\mathbf{m_{ice} = 0.14 \; kg}$

This the value of ice in kilograms which required to drop in to the water to make the final temperature of the system 29.9 ^oC.