Question

Suppose that we replace the aluminum with a mystery metal and repeat the experiment in the video. As in the video, the mass of the metal is the same as that of the water. Room temperature is about 20∘C before the start of the experiment. The water heats up to 40∘C, and the mystery metal heats up to 80∘C. Compared to that of water, the heat capacity of our mystery metal isa) three times greater.b) one-third as great.c) half as great.d) two times greater.e) the same.

Answer 1

The heat capacity of our mystery metal is b) one-third as great

Further explanation

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

the mass of the metal is the same as that of the water

so m water = m metal

the heat energy used to heat water and metals is the same, so

Q water = Q metal

$\displaystyle m_mc_m (T-Tm) = m_wc_w (T-Tw)$

m = metal

w = water

T = the final temperature of the mixture

Tm = metal temperature

Tw = water temperature

so the equation becomes:

$\displaystyle c_m(80-20)=c_w(40-20)\\\\60c_m=20c_w\\\\\frac{c_m}{c_w}=\frac{20}{60}=\frac{1}{3}$

Learn more

the difference between temperature and heat

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Specific heat

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relationships among temperature, heat, and thermal energy.

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When heat is added to a substance

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Keywords: heat, temperature,the heat capacity

Answer 2

Answer:

b) one-third as great.

Explanation:

As we know that same heat is supplied in this experiment

so we will have

$Q = ms\Delta T$

now we know that both are initially at same temperature

then their final temperatures are

$T_1 = 40 degree$

$T_2 = 80 degree$

now we have

$m_1s_1 (40 - 20) = m_2s_2(80 - 20)$

so we have

$m_2s_2 = \frac{20}{60} m_1s_1$

so heat capacity of mystery metal is 1/3 times that of water