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3 years ago

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How much heat do you need to raise the temperature of 150 g of ice from -30°C to -15°c?

1 answer:

OLga [1]3 years ago

4 0

To solve this question, you must use the formula: q=mc(change in temperature), where q is heat, m is mass, C is specific heat and temperature change is temperature change. The specific heat for ice is 2.1kJ/Kg x K (given). The change in temperature is 15 degrees Celsius (which you should change to kelvins so you can cancel out units), or 273 + 15 = 288K. The mass is 150 grams, which is 0.15 kg. Now, we can solve for q, heat. We will do this by substituting variables into the formula. After simplifying and cancelling out units, the answer we get is: 90.72kJ.

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Explanation:

To calculate the initial temperature of the water:

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = $5.73\times 10^5J$

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 2230 g

$T_{final}$ = final temperature of water = $100^0C$

$T_{initial}$ = initial temperature of metal = ?

Now put all the given values in the above formula, we get:

$5.73\times 10^5J=2230g\times 4.18J/g^oC\times (100-T_i)^0C$

$(100-T_i)=61.5$

$T_i=100-61.5)=38.5^0C$

Thus, the initial temperature of the water is $38.5^0C$

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