The addition of 9.0×105 J is required to convert a block of ice at -10 ∘C to water at 11 ∘C. i need help this is due in less tha

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1 answer:

Law Incorporation [45] · Answer 1 · 2022-05-13T08:57:29+03:00

The mass of the block of ice is 2.24 kg

Explanation:

The amount of heat needed for the whole process consists of three different amounts of heat:

$Q_1$ : the amount of heat needed to raise the temperature of the block of ice from $-10^{\circ}C$ to $0^{\circ}C$

$Q_2$ : the amount of heat needed to melt the block of ice at melting point

$Q_3$ : the amount of heat needed to raise the temperature of the water from $0^{\circ}C$ to $11^{\circ}C$

The total amount of heat needed can be written as

$Q=Q_1+Q_2+Q_3=mC_i\Delta T_1 + m\lambda_f + mC_w\Delta T_3$

where we have:

$Q=9.0 \cdot 10^5 J$ (total amount of heat required)

m is the mass of the block of ice

$C_i = 2108 J/kg^{\circ}C$ is the specific heat of ice

$\lambda_f=3.34\cdot 10^5 J/kg$ is the latent heat of fusion of ice

$C_w=4186 J/kg^{\circ}C$ is the specific heat capacity of water

$\Delta T_1 = 0-(-10)=10^{\circ}C$ is the change in temperature in the 1st process

$\Delta T_3 = 11-0=11^{\circ}C$ is the change in temperature in the 3rd process

Solving the equation for m, we find the mass of the block of ice:

$m=\frac{Q}{C_i\Delta T_1 + \lambda_f+C_w\Delta T_3}=\frac{9.0\cdot 10^5}{(2108)(10)+3.34\cdot 10^5+(4186)(11)}=2.24 kg$

Learn more about specific heat capacity:

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