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Vanyuwa [196]
3 years ago
15

A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of water is 4.18 J•g^-1•C^-1. By how much will it’s temp

erature rise, assuming it does not melt?
Chemistry
1 answer:
MrMuchimi3 years ago
0 0

Answer: The temperature rise is 0.53^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed by ice = 5280 J

m = mass of ice = 2.40 kg = 2400 g   (1kg=1000g)

c = heat capacity of water = 4.18J/g^0C

Initial temperature  = T_i

Final temperature = T_f  

Change in temperature ,\Delta T=T_f-T_i=?

Putting in the values, we get:

5280J=2400g\times 4.18J/g^0C\times \Delta T

\Delta T=0.53^0C

Thus the temperature rise is 0.53^0C

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\begin{array}{rcl}2 \, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\2 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{2 \times 250\text{ mL}}{100}\\\\ & = & \textbf{5 mL}\\\end{array}

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