91Ó°ÊÓ

a) The following is the heat equation for calculating energy:

Heat $=mass\left(m\right)×$temperature change$\left(\mathrm{Δ}T\right)×$specific heat $(SH)$

Mass of the water is $85\mathrm{g}$.

The temperature fluctuates between$45°\mathrm{C}$to $25°\mathrm{C}$

$\mathrm{Δ}T={45}^{∘}\mathrm{C}−{25}^{∘}\mathrm{C}$

$=20°C$

Specific heat of water $=1\mathrm{cal}/{\mathrm{g}}^0\mathrm{C}$

Calculate the heat lost as follows:

Heat lost

$=85\mathrm{g}×20°\mathrm{C}×1\mathrm{cal}/\mathrm{g}°\mathrm{C}$

As a result, the heat is$1700Cal$

(b)

The following is the heat equation for calculating energy:

Heat $=mass\left(m\right)×$temperature change $(∆T)×$specific heat $\left(SH\right)$

Mass of the water is $75\mathrm{g}$.

The temperature fluctuates between$22°C$to $66°C$

$\mathrm{Δ}T={66}^{∘}\mathrm{C}−{22}^{∘}\mathrm{C}$

$={44}^{∘}\mathrm{C}$

Specific heat of water $=4.184\mathrm{J}/{\mathrm{g}}^0\mathrm{C}$

Calculate the heat as follows:

$\text{Heat}=75\mathrm{g}×44°\mathrm{C}×4.184\mathrm{J}/\mathrm{g}°\mathrm{C}$

$=13807J$

As a result, the heat is$=13807J$

(C)

The following is the heat equation for calculating energy:

Heat $=mass\left(m\right)×$temperature change $\left(\mathrm{Δ}T\right)×$specific heat $\left(SH\right)$

Mass of the water is $5.0\mathrm{kg}$

The temperature fluctuates between $22°C$to $28°C$

$\mathrm{Δ}T={28}^{∘}\mathrm{C}−{22}^{∘}\mathrm{C}$

$=6°C$

Specific heat of water $=1\mathrm{cal}/{\mathrm{g}}^0\mathrm{C}$

Calculate the heat as follows:

$\text{Heat}=5000\mathrm{g}×6°\mathrm{C}×1\mathrm{cal}/\mathrm{g}{}^{∘}\mathrm{C}$

$=30000cal$

To calculate the heat, do the following:

$30000\mathrm{cal}×\frac{1\mathrm{kcal}}{1000\mathrm{cal}}=30\mathrm{kca}l$

As a result, the heat is$3000cal$

d)The following is the heat equation for calculating energy:

Heat$=mass\left(m\right)×$temperature change$(∆T)×$ specific heat$\left(SH\right)$

Mass of the water is $224\mathrm{g}$.

The temperature fluctuates between $18°C$to $185°C$

$\mathrm{Δ}T={185}^{∘}\mathrm{C}−{18}^{∘}\mathrm{C}$

$={167}^{∘}\mathrm{C}$

Specific heat of gold $=0.129\mathrm{J}/{\mathrm{g}}^0\mathrm{C}$

To calculate the heat, do the following:

$=4.83×{10}^3\mathrm{J}$

Convert the calories to kilocalories

$4.83×{10}^3\mathrm{J}×\frac{1\mathrm{kJ}}{1000\mathrm{J}}=4.83\mathrm{kJ}$

As a result, the heat is$=4.83×{10}^3\mathrm{J}$