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2 years ago

60 grams of ice will require _____ calories to raise the temperature 1°C.

Chemistry

1 answer:

guajiro [1.7K]2 years ago

7 0

Answer:

60 grams of ice will require 30.26 calories to raise the temperature 1°C.

Explanation:

The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:

Q = m.c.ΔT,

where, Q is the amount of heat released or absorbed by the system.

m is the mass of the ice (m = 60.0 g).

c is the specific heat capacity of ice (c = 2.108 J/g.°C).

ΔT is the temperature difference (ΔT = 1.0 °C).

∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.

It is known that 1.0 cal = 4.18 J.

∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.

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When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t

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The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

$Mn(s)+2HCl(aq)\rightarrow MnCl_2(aq)+H_2(g)$

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

$m=1.00 g/mL\times 100.0 mL = 100 g$

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

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

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{final}$ = final temperature = $23.1^oC$

$T_{initial}$ = initial temperature = $28.9^oC$

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

$q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC$

$q=2,242.4 J=2.242 kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = $\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol$

$\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol$

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

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